Kiritow/OJ-Problems-Source
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create by jzf@2018/06/20

[TOC]

作业

627. 交换工资 @2018/06/20

jzf :517ms

update salary 
    set sex = case when sex ='m' then 'f' else 'm' end;

kiritow:563ms

    update salary set sex = case sex
    when 'f' then 'm'
    when 'm' then 'f'
    end

3. 无重复字符的最长子串 @2018/06/20

北河

* 第一版 :392ms
```c++
class Solution {
public:
    int lengthOfLongestSubstring(string s) {
        map<char, short> v;
        int max_num = 0;
        for (int i = 0; i < s.size(); ++i) {
            if (v.find(s[i]) != v.end()) {
                if (max_num < v.size()) {
                    max_num = v.size();
                }
                i = v.find(s[i])->second + 1;
                v.clear();
            }
            if (i >= s.size()) {
                break;
            }
            v[s[i]] = i;
            if (max_num < v.size()) {
                max_num = v.size();
            }
        }
        return max_num;
    }
};

```
* 第2版 :32ms
```c++
class Solution {
public:
    int lengthOfLongestSubstring(string s) {
        map<char, short> v;
        int max_num = 0;
        for (int i = 0; i < s.size(); ++i) {
            if (v.find(s[i]) != v.end()) {
                if (max_num < v.size()) {
                    max_num = v.size();
                }
                ++i;
                while (i < s.size()) {
                    if (v.find(s[i]) != v.end()) {
                        ++i;
                        continue;
                    }
                    --i;
                    break;
                }
                if (i >= s.size()) {
                    return max_num;
                }
                i = v.find(s[i])->second + 1;
                v.clear();
            }
            if (i >= s.size()) {
                break;
            }
            v[s[i]] = i;
            if (max_num < v.size()) {
                max_num = v.size();
            }
        }
        return max_num;
    }
};
```

jzf :252 ms 8

class Solution {
public:
    int lengthOfLongestSubstring(string s) {
        int num =1;
        int max =0;
        if(s.length()==1)
        {
            return 1;
        }
        for(size_t i =0;i<s.length();i++)
        {
            
            for(size_t j=i+1;j<s.length();j++)
            {
                int flag =0;
                for(size_t k =i;k<j;k++)
                {
                    if(s[j]==s[k])   
                    {
                        flag =1;
                        break;
                    }
                    num++;
                }
                    if(num>max)
                    {
                        max =num;
                    }
                if(flag ==1)
                {
                    break;
                }
                num =1;
            }
             num =1;
        }
        return max;
    }
};

puck :544ms 2

#ifndef SOLUTION_H
#define SOLUTION_H

#include <algorithm>
#include <string>
#include <set>
using std::string;
class Solution
{
  public:
    int lengthOfLongestSubstring(string s)
    {
        int result{};
        for (std::size_t i{}; i < s.size(); ++i)
        {
            string tmp_str = s.substr(i);
            int cu_length{};
            std::set<char> set_except;
            
            for (auto ch : tmp_str)
            {
                if (set_except.end() == set_except.find(ch))
                {
                    set_except.insert(ch);
                    ++cu_length;
                }
                else
                {
                    break;
                }
            }

            result = std::max(result, cu_length);
        }

        return result;
    }
};

#endif

kiritow ：24ms 2

#include <cstdlib>
#include <cstring>
class Solution {
public:
    int lengthOfLongestSubstring(string s) {
        int sz = s.size();
        int bin[256];
        int maxlen = 0;
        int clen = 0;
        int L = 0;
        while (L < sz)
        {
            memset(bin, 0, sizeof(int) * 256);
            int i;
            for (i = L; i < sz; i++)
            {
                if (bin[s[i]])
                {
                    maxlen = clen > maxlen ? clen : maxlen;
                    clen = 0;
                    L++;
                    break;
                }
                else
                {
                    bin[s[i]] = 1;
                    clen++;
                }
            }
            if (i == sz)
            {
                maxlen = clen > maxlen ? clen : maxlen;
                break;
            }
        }
        return maxlen;
    }
};

知识点

桶排序
sync_with_stdio

总结

1, 对比代码发现，利用现有的数据结构，可以快速出活，而且代码简洁，逻辑清晰。

2, STL不好好用，很慢。

494. 目标和 @2018/06/21

jzf

* 第一版 超时  复杂度 n*2^n
```c++
class Solution {
public:  int findTargetSumWays(vector<int>& nums, int S) {
    int n = 0;
    int sum = 0;
    int b = 0;
    int size =nums.size();
    for (size_t k = 0; k <size ; k++)
    {
        b = ((1 << k) | b);
    }
    for (size_t i = 0; i <= b; i++)
    {
        n = 0;
        for (size_t j = 0; j < size; j++)
        {
            n += ((1 << j)&i) ? -nums[j] : nums[j];
        }
        if (n == S)
        {
            sum++;
        }
    }
    return sum;
    }
};
```
* 第二版

kiritow ：24ms 2

class Solution {
public:
    int findTargetSumWays(vector<int>& nums, int S) {
        if (S < -1000 || S>1000) return 0;

        int sz = nums.size();
        int msz = sizeof(int) * 2048;
        int _bin[2048] = { 0 };
        int _xbin[2048] = { 0 };
        int* p = _bin;
        int* q = _xbin;
        memset(p, 0, msz);
        p[1000] = 1;
        
        for (int i = 0; i < sz; i++)
        {
            memset(q, 0, msz);
            for (int j = 0; j < 2048; j++)
            {
                if (p[j])
                {
                    //printf("p[%d]=%d\n", p[j]);
                    // now=j-1000;
                    int L = j - nums[i]; // now-nums[i]+1000
                    int R = j + nums[i];
                    q[L] += p[j];
                    q[R] += p[j];
                    //printf("q[%d]=%d\n", L, q[L]);
                    //printf("q[%d]=%d\n", R, q[R]);
                }
            }
            std::swap(p, q);
        }
        return p[S + 1000];
    }
};

知识点

dfs

bfs

剪枝

记忆搜索

去掉重复

没写dp，身败名裂--惠惠语录

601. 体育馆的人流量@2018/06/21

jzf :521ms 抄

    select s1.* 
    from stadium  as s1, stadium  as s2,stadium  as s3 
    where (((s2.id =s1.id+1) and (s3.id =s1.id+2))
           or((s2.id =s1.id-1) and (s3.id =s1.id+1))
           or((s2.id =s1.id-2) and (s3.id =s1.id-1)))
        and 
         (s1.people>=100 and s2.people>=100 and s3.people >=100)
    group by s1.id

知识点

mysql必知必会 16章 自连接

总结

1，看了下最快的sql语句，感叹逻辑差不过的东西，差距真大呀

9. 回文数@2018/06/22

羽柔子 372ms

var isPalindrome = function(x) {
    return x.toString()===x.toString().split("").reverse().join("");
};
  • 第二版 ：128ms
class Solution {
    int list[10],len=0,i=0;
public:
    bool isPalindrome(int x) {
        if(x<0)return false;
        len=i=0;
        //分解
        while(true){
            if(x>=10){
                list[len++]=x%10;
                x/=10;
            }
            else{
                list[len++]=x;
                break;
            }
        }

        //判断
        --len;
        for(i=0;i<len&&list[i]==list[len];i++,len--);
        return i>=len;
    }
};

jzf

  • 第一版 ：232ms 
    class Solution {
public:
    bool isPalindrome(int x) {
        int mod;
        int sh =x;
        vector<int> vec;
        if(x <0)
        {
            return false;
        }
        if(x ==0)
        {
            return true;
        }

        for(size_t i =0;sh>0;i++)
        {
            vec.push_back(sh%10);
            sh = sh/10;
        }
        vector<int>::iterator i=vec.begin();
        vector<int>::iterator j=vec.end()-1;
        for(;i<j;i++,j--)
        {
            if(*i!=*j)
            {
                return false;
            }
        }
        return true;  
    }
};
  • 第2版 ：140ms 
    class Solution {
public: bool isPalindrome(int x) {
       int mod;
       int sh =x;
       vector<int> vec;
       if(x <0)
       {
           return false;
       }
       if(x ==0)
       {
           return true;
       }
       if(x%10 ==0)
       {
           return false;
       }
       for(size_t i =0;sh>0;i++)
       {
           vec.push_back(sh%10);
           sh = sh/10;
       }
       vector<int>::iterator i=vec.begin();
       vector<int>::iterator j=vec.end()-1;
       for(;i<j;i++,j--)
       {
           if(*i!=*j)
           {
               return false;
           }
       }
       return true;  
   }
};

北河 244ms

  • 第一版
class Solution {
public:
    bool isPalindrome(int x) {
        char p[10], q[10];
        short size = sprintf(p, "%d", x);
        reverse_copy(p, p + size, q);
        return x < 0 ? false : (mismatch(p, p + size / 2 + 1, q).first == (p + size / 2 + 1) ? true : false);               
    }
};
  • 第二版
class Solution {
public:
    bool isPalindrome(int x) {
        if (x < 0) {
            return false;
        }
        string s = to_string(x);
        return mismatch(s.begin(), s.end(), s.rbegin()).first == s.end();
    }
};

kiritow 120ms

#include <cstdio>
class Solution {
public:
    bool isPalindrome(int x) {
        /*
        // Converting to string is too slow!!
        char buff[64] = { 0 };
        sprintf(buff, "%d", x);
        int len = strlen(buff);
        int L = 0, R = len - 1;
        while (L < R)
        {
            if (buff[L] != buff[R]) return false;
            ++L;
            --R;
        }
        return true;
        */
        if (x < 0) return false;
        int buff[64] = { 0 };
        int c = 0;
        int tx = x;
        while (tx > 0)
        {
            buff[c++] = tx % 10;
            tx /= 10;
        }
        int L = 0, R = c - 1;
        while (L < R)
        {
            if (buff[L] != buff[R]) return false;
            ++L;
            --R;
        }
        return true;
    }
};

puck

#include <string>

class Solution
{
  public:
    bool isPalindrome(int x)
    {
        if (x < 0)
            return false;

        std::string tmp = std::to_string(x);

        std::size_t bi_size = tmp.size() / 2;
        auto half = tmp.begin() + bi_size;
        auto iter = tmp.begin();
        auto riter = tmp.rbegin();
        for (; iter != half; ++iter, ++riter)
        {
            if (*iter != *riter)
                return false;
        }
        return true;
    }
};

总结

1， 我在想一个问题 羽柔子撤回了一条消息 喵在想一个问题 --羽柔子语录

225. 用队列实现栈@2018/06/22

羽柔子 0ms

class MyStack {
    int stack[10],index=0;
    
public:
    /** Initialize your data structure here. */
    MyStack() {
    }
    
    /** Push element x onto stack. */
    void push(int x) {
        stack[index++]=x;
    }
    
    /** Removes the element on top of the stack and returns that element. */
    int pop() {
        return stack[--index];
    }
    
    /** Get the top element. */
    int top() {
        return stack[index-1];
    }
    
    /** Returns whether the stack is empty. */
    bool empty() {
        return !index;
    }
};

kiritow

#include <list>
class MyStack {
public:
    /** Initialize your data structure here. */
    MyStack() {
        
    }

    /** Push element x onto stack. */
    void push(int x) {
        lst.push_back(x);
    }

    /** Removes the element on top of the stack and returns that element. */
    int pop() {
        int t = lst.back();
        lst.pop_back();
        return t;
    }

    /** Get the top element. */
    int top() {
        return lst.back();
    }

    /** Returns whether the stack is empty. */
    bool empty() {
        return lst.empty();
    }
private:
    std::list<int> lst;
};

28. 实现strStr() @2018/06/23

jzf

  • 第一版 4ms
class Solution {
public:
    int strStr(string haystack, string needle) {
        string::iterator ihay;
        string::iterator inee;
        int ni =haystack.size();
        int nn =needle.size();
        int ret = -1;
        
        if(nn==0)
        {
            return 0;
        }
        if(ni<nn)
        {
            return -1;
        }
        for(size_t i=0;i<ni-nn+1;i++)
        {
            if(!(haystack.substr(i,nn)).compare(needle))
            {
                ret =i;
                break;
            }
        }
        
        return ret;
    }
};

羽柔子

class Solution {
public:
    int strStr(string haystack, string needle) {
        if(needle.size()==0)return 0;
        if(haystack.size()<needle.size())return -1;
        size_t 
            i,j,
            ilen=haystack.size()-needle.size(),
            jlen=needle.size(); 
        for( i = 0; i <= ilen; i++){
            for( j = 0; j < jlen; j++)
                if(haystack[i+j]!=needle[j])
                    goto A;
            return i;
            A:;
        }
        return -1;
    }
};

puck

class Solution
{
  public:
    int strStr(string haystack, string needle)
    {
        std::size_t size_a{haystack.size()};
        std::size_t size_b{needle.size()};

        if (0 == size_b)
        {
            return 0;
        }

        bool flag{};

        for (int i{}; i < size_a; ++i)
        {
            if (haystack.at(i) == needle.at(0))
            {
                flag = true;

                for (int j{1}; j < size_b; ++j)
                {
                    if (i + j >= size_a || haystack.at(i + j) != needle.at(j))
                    {
                        flag = false;
                        break;
                    }
                }

                if (flag)
                {
                    return i;
                }
            }
        }

        return -1;
    }
};

823. Binary Trees With Factors@2018/06/23

196. 删除重复的电子邮箱@2018/06/23

北河

delete from Person where id in (select c.id from (select a.id from Person a  join Person b on a.Email = b.Email and a.Id > b.Id) c);

48. 旋转图像@2018/06/24

jzf

  • 4ms
class Solution {
public:
    void rotate(vector<vector<int>>& matrix) {
        
        int tmp;
        for(size_t i=0; i<matrix.size();i++)
        {
            for(size_t j=i+1;j<matrix.size();j++)
            {
                
                tmp =matrix[i][j];
                matrix[i][j] =matrix[j][i];
                matrix[j][i] =tmp;
            }
        }
        for(size_t i =0;i<matrix.size();i++)
        {
            reverse(matrix[i].begin(),matrix[i].end());
        }
    }
};

kiritow

class Solution {
public:
    void rotate(vector<vector<int>>& matrix) {
        int n = matrix.size();
        if (n % 2) // odd number, easy
        {
            int mid = n / 2;

            for (int dis = 1; dis <= mid; dis++)
            {
                // do chain swap at distance:dis
                int swap_n = 2 * dis + 1;
                // save right col
                vector<int> tvec;
                int begin_pos = mid - dis;
                for (int i = 0; i < swap_n; i++)
                {
                    tvec.push_back(matrix[begin_pos + i][mid + dis]);
                }
                // start rotate
                // RIGHT <- UPPER
                for (int i = 0; i < swap_n; i++)
                {
                    //printf("matrix[%d][%d]<-matrix[%d][%d]\n", begin_pos + i, mid + dis, mid - dis, begin_pos + i);
                    matrix[begin_pos + i][mid + dis] = matrix[mid - dis][begin_pos + i];
                }
                // UPPER <- LEFT
                for (int i = 0; i < swap_n; i++)
                {
                    //printf("matrix[%d][%d]<-matrix[%d][%d]\n", mid - dis, begin_pos + i, begin_pos + i, mid - dis);
                    matrix[mid - dis][begin_pos + (swap_n - i - 1)] = matrix[begin_pos + i][mid - dis];
                }
                // LEFT <- DOWNER
                for (int i = 0; i < swap_n; i++)
                {
                    //printf("matrix[%d][%d]<-matrix[%d][%d]\n", begin_pos + i, mid - dis, mid + dis, begin_pos + i);
                    matrix[begin_pos + i][mid - dis] = matrix[mid + dis][begin_pos + i];
                }
                // DOWNER <- TEMP
                for (int i = 0; i < swap_n; i++)
                {
                    //printf("matrix[%d][%d]<-tvec[%d]\n", mid + dis, begin_pos + i, i);
                    matrix[mid + dis][begin_pos + (swap_n - i - 1)] = tvec[i];
                }

                // debug
                //print(matrix);
            }
        }
        else // oh, this stuff sucks.
        {
            int mid = n / 2 ;

            for (int dis = 1; dis <= mid ; dis++)
            {
                // do chain swap at distance:dis
                int swap_n = 2 * dis;
                // save right col
                vector<int> tvec;
                int begin_pos = mid - dis;
                for (int i = 0; i < swap_n; i++)
                {
                    tvec.push_back(matrix[begin_pos + i][mid + dis - 1]);
                }
                // start rotate
                // RIGHT <- UPPER
                for (int i = 0; i < swap_n; i++)
                {
                    //printf("matrix[%d][%d]<-matrix[%d][%d]\n", begin_pos + i, mid + dis, mid - dis, begin_pos + i);
                    matrix[begin_pos + i][mid + dis - 1] = matrix[mid - dis][begin_pos + i];
                }
                // UPPER <- LEFT
                for (int i = 0; i < swap_n; i++)
                {
                    //printf("matrix[%d][%d]<-matrix[%d][%d]\n", mid - dis, begin_pos + i, begin_pos + i, mid - dis);
                    matrix[mid - dis][begin_pos + (swap_n - i - 1)] = matrix[begin_pos + i][mid - dis];
                }
                // LEFT <- DOWNER
                for (int i = 0; i < swap_n; i++)
                {
                    //printf("matrix[%d][%d]<-matrix[%d][%d]\n", begin_pos + i, mid - dis, mid + dis, begin_pos + i);
                    matrix[begin_pos + i][mid - dis] = matrix[mid + dis -1][begin_pos + i];
                }
                // DOWNER <- TEMP
                for (int i = 0; i < swap_n; i++)
                {
                    //printf("matrix[%d][%d]<-tvec[%d]\n", mid + dis, begin_pos + i, i);
                    matrix[mid + dis -1][begin_pos + (swap_n - i - 1)] = tvec[i];
                }

                // debug
                //print(matrix);
            }
        }
    }
};

puck

#include <algorithm>

class Solution
{
  public:
    void rotate(vector<vector<int>> &matrix)
    {
        auto n = matrix.size();
        auto max = n - 1;
        auto half = n / 2;

        for (int i{}; i < half; ++i)
        {
            for (int j{}; j < half; ++j)
            {
                std::swap(matrix[i][j], matrix[j][max - i]);
                std::swap(matrix[i][j], matrix[max - i][max - j]);
                std::swap(matrix[i][j], matrix[max - j][i]);
            }
        }

        if (n == 2 * half + 1)
        {
            for (int i{}; i < half; ++i)
            {
                std::swap(matrix[i][half], matrix[half][max - i]);
                std::swap(matrix[i][half], matrix[max - i][half]);
                std::swap(matrix[i][half], matrix[half][i]);
            }
        }
    }
};

4. 两个排序数组的中位数@2018/06/24

种子 似乎不符合复杂度限制

class Solution {
public:
    double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
        copy(nums2.begin(),nums2.end(),back_inserter(nums1));
        sort(nums1.begin(),nums1.end());
        return nums1.size()%2?\
            double(nums1[nums1.size()/2]):\
            double(nums1[nums1.size()/2]+nums1[nums1.size()/2-1])/2;
    }
};

北河 48ms

class Solution {
public:
    double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
        std::multiset<int> v;
        copy(nums1.begin(), nums1.end(), inserter(v, v.begin()));
        copy(nums2.begin(), nums2.end(), inserter(v, v.begin()));
        std::multiset<int>::iterator p, q;
        p = q = v.begin();
        return v.size() % 2 == 0 ? ((double)*(advance(p, v.size() / 2 - 1), p) + (double)*(advance(q, v.size() / 2), q))/ 2 : (double)*(advance(p, v.size() / 2), p);
    }
};

kiritow

#include <algorithm>
class Solution {
public:
    void forward_until(vector<int>& nums1, vector<int>& nums2,
        int lenA,int lenB,
        int& idxA, int& idxB, int& passed, int midLen)
    {
        while (idxA < lenA&&idxB < lenB && passed < midLen)
        {
            if (nums1[idxA] < nums2[idxB])
            {
                idxA++;
                passed++;
            }
            else
            {
                idxB++;
                passed++;
            }
        }
        while (idxA < lenA&&passed < midLen)
        {
            idxA++;
            passed++;
        }
        while (idxB < lenB&&passed < midLen)
        {
            idxB++;
            passed++;
        }
    }

    int getCurrent(vector<int>& nums1, vector<int>& nums2, 
        int lenA,int lenB,
        int idxA, int idxB)
    {
        if (idxA < lenA&&idxB < lenB)
        {
            return std::min(nums1[idxA], nums2[idxB]);
        }
        else if (idxA < lenA)
        {
            return nums1[idxA];
        }
        else
        {
            return nums2[idxB];
        }
    }

    double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
        int lenA = nums1.size();
        int lenB = nums2.size();
        int sumLen = lenA + lenB;
        int midLen = sumLen / 2 - (1 - (sumLen % 2));
        int idxA = 0;
        int idxB = 0;
        int passed = 0;
        
        forward_until(nums1, nums2, lenA, lenB, idxA, idxB, passed, midLen);

        if (sumLen % 2)
        {
            return getCurrent(nums1, nums2, lenA, lenB, idxA, idxB);
        }
        else
        {
            int step1 = getCurrent(nums1, nums2, lenA, lenB, idxA, idxB);
            
            forward_until(nums1, nums2, lenA, lenB, idxA, idxB, passed, midLen + 1);
            
            int step2 = getCurrent(nums1, nums2, lenA, lenB, idxA, idxB);

            return (step1 + step2) / 2.0;
        }
    }
};

jzf

算了，我就算是抄上边几个的吧。

小白

  • 第一版 40ms

class Solution {
public:

    double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
            vector<int> ab;

	std::merge(nums1.begin(), nums1.end(), nums2.begin(), nums2.end(), std::back_inserter(ab));

	return (double(ab[(ab.size() >> 1) - !(ab.size() & 1)]) + double(ab[(ab.size() >> 1)])) *0.5;
    }
};
  • 第二版
class Solution {
public:

    double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
            vector<int> ab;

	std::merge(nums1.begin(), nums1.end(), nums2.begin(), nums2.end(), std::back_inserter(ab));

	return (double(ab[(ab.size() >> 1) - !(ab.size() & 1)]) + double(ab[(ab.size() >> 1)])) *0.5;
    }
};
  • 第三版 40ms

class Solution {

public:

    double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {

        vector<int> ab;

        ab.reserve(nums1.size() + nums2.size());

	std::merge(nums1.begin(), nums1.end(), nums2.begin(), nums2.end(), std::back_inserter(ab));

	return double(ab[(ab.size() >> 1) - !(ab.size() & 1)] + ab[(ab.size() >> 1)]) * 0.5;

    }

};

总结

1，北河和种子的体现了使用STL的巨大优势，利用现有的工具，确实好很多。

而我在解题的过程中出现了自创轮子没本事放弃，和不熟悉STL等问题

2，算法还是要学的，STL的复杂度也要会分析

知识点

1， std::merge 2，copy

724. 寻找数组的中心索引@2018/06/25

jzf 28ms

class Solution {
public:
    int pivotIndex(vector<int>& nums) {
        vector<int>::iterator iter;
        int sum =0;
        int ret =-1;
        for(iter =nums.begin();iter!=nums.end();iter++)
        {
            sum+=*iter;
        }
        
        int tmp =0;
        
        for(size_t i=0;i<nums.size();i++)
        {
            if((sum -nums[i]-tmp)==tmp)
            {
                ret =i;
                break;
            }
            tmp+=nums[i];
        }
        return ret;
    }
};

羽柔子

var pivotIndex = function(nums) {
    var r = 0,l=0,len = nums.length;
    if(len<2)return -1;
    for(
        var i = 1;
        i < len;
        i++
    )r += nums[i];
    for(
        var i=0;
        l!=r && i<len;
    ){
        l+=nums[i];
        i++;
        r-=nums[i];
    }
    return i>=len?-1:i;
};


class Solution {
public:
    int pivotIndex(vector<int>& nums) {
        int  r = 0,l=0,len = nums.size(),i;
        if(len<2)return -1;
    for( i = 1; i < len; i++ )r += nums[i];
    for(i=0; l!=r && i<len;){
        l+=nums[i];
        i++;
        r-=nums[i];
    }
    return i>=len?-1:i;
    }
};

puck

#include <algorithm>

class Solution
{
public:
    int pivotIndex(vector<int> &nums)
    {
        int n{-1};
        auto iter = std::find_if(nums.begin(), nums.end(), [&](const int a) {
            return ++n, std::accumulate(nums.begin(), nums.begin() + n, 0) ==
                            std::accumulate(nums.begin() + n + 1, nums.end(), 0);
        });

        if (iter == nums.end())
        {
            return -1;
        }
        return iter - nums.begin();
    }
};

小白

5. 最长回文子串@2018/06/25

jzf

  • 第一版 超时
class Solution {
public:
    string longestPalindrome(string s) {
        int len =s.size();
        string retStr="";
        for(size_t i = len; i>0;i--)
        {
            for(size_t j =0;j+i<=len;j++)
            {
                    string str =s.substr(j,i);
                    string str2 =str;
                    reverse(str2.begin(),str2.end());
                    
                    if(!str.compare(str2))
                    {
                        retStr =str;
                        return retStr;
                    }

            }
        }
        return retStr;
    }
};
  • 第二版 超时
class Solution {
public:
    string longestPalindrome(string s) {
        int len =s.size();
        string retStr=s.substr(0,1);
        for(size_t i = len; i>1;i--)
        {
            for(size_t j =0;j+i<=len;j++)
            {
                    string str =s.substr(j,i);
                    string str2 =str;
                    reverse(str2.begin(),str2.end());
                    
                    if(!str.compare(str2))
                    {
                        retStr =str;
                        return retStr;
                    }

            }
        }
        return retStr;
    }
};
  • 8ms
class Solution {
public:
int  func(string &str, int n)
{
	int ret = 1;
	size_t k = 1;
	for ( ;k <= n && (n + k)<=str.size(); k++)
	{
		if (str[n - k] == str[n + k + 1])
		{
			ret ++;
		}
		else
		{
			break;
		}
	}
	return ret;
}
int  func2(string &str, int n)
{
	int ret = 1;
	size_t k = 1;
	for (; k <= n && (n + k+3) <= str.size(); k++)
	{
		if (str[n - k] == str[n + k+2])
		{
			ret++;
		}
		else
		{
			break;
		}
	}
	return ret;
}
string longestPalindrome(string s) {
	int len = s.size();
	int max = 0;
	int center = 0;
	int f = 0;
	if (len<2)
	{
		return s;
	}
	if (len == 2)
	{
		if (s[0] == s[1])
		{
			return s;
		}
		else
		{
			return s.substr(0, 1);
		}
	}

	std::vector<int> vec1;
	for (size_t i = 0; i<len - 1; i++)
	{
		if (s[i] == s[i + 1])
		{
			vec1.push_back(i);
		}
	}

	for (size_t i = 0; i<vec1.size(); i++)
	{
		int tmp = func(s, vec1[i]);
		if (tmp>max)
		{
			max = tmp;
			center = vec1[i];
			f = 1;
		}
		if (max + i>s.size())
		{
			break;
		}
	}
	std::vector<int> vec2;
	for (size_t i = 0; i<len - 2; i++)
	{
		if (s[i] == s[i + 2])
		{
			vec2.push_back(i);
		}
	}
	for (size_t i = 0; i<vec2.size(); i++)
	{
		int tmp = func2(s, vec2[i]);
		if (tmp + 1>max)
		{
			max = tmp ;
			center = vec2[i];
			f = 2;
		}
		if (max + i>s.size())
		{
			break;
		}
	}
	if (f == 0)
	{
		return s.substr(0,1);
	}
	if (f == 1)
	{
		return s.substr(center - max+1, max * 2);
	}
	if (f == 2)
	{
		return s.substr(center -max+1, max * 2 + 1);
	}
}
};

小白

  • 第一版 超时
std::pair<string::const_iterator, string::const_iterator> longestPalindrome(string::const_iterator a, string::const_iterator b)

{

	if (a==b || std::mismatch(a, b, make_reverse_iterator(b)).first == b)

		return std::make_pair(a, b);

	auto str1 = longestPalindrome(a + 1, b), str2 = longestPalindrome(a, b - 1);

	return (str1.second - str1.first) > (str2.second - str2.first) ? str1 : str2;

}

string longestPalindrome(const string &s) {

	auto pr = longestPalindrome(s.begin(), s.end());

	return std::string(pr.first, pr.second);

}


class Solution {
public:
    string longestPalindrome(string s) {
    auto lr_max = std::make_pair( s.crend(), s.cbegin() );
	auto lr_dist_comp = [](decltype(lr_max) a, decltype(lr_max) b) 
    {return std::distance(a.first.base(), a.second) < 
        std::distance(b.first.base(), b.second); };
	for (auto cent = lr_max; cent != std::make_pair(s.crbegin(), s.cend());             std::distance(cent.first.base(), cent.second) == 0 ? ++cent.second : (--cent.first).base() )
		lr_max = std::max(lr_max, std::mismatch((cent.first - 1), s.crend(), cent.second), lr_dist_comp);
	return std::string(lr_max.first.base(), lr_max.second);
    }
};

384. 打乱数组@2018/06/26

jzf

class Solution {
    vector<int> nums;
public:
    Solution(vector<int> nums) {
        this->nums = nums;
    }
    
    /** Resets the array to its original configuration and return it. */
    vector<int> reset() {
        return nums;
    }
    
    /** Returns a random shuffling of the array. */
    vector<int> shuffle() {
        vector<int> result(nums);
        for (int i = 0;i < result.size();i++) {
            int pos = rand()%(result.size()-i);
            swap(result[i+pos], result[i]);
        }
        return result;
    }
};

北河

class Solution {
public:
    Solution(vector<int> nums) :m_nums(nums.begin(), nums.end()) {
    }
    
    /** Resets the array to its original configuration and return it. */
    vector<int> reset() {
        return m_nums;
    }
    
    /** Returns a random shuffling of the array. */
    vector<int> shuffle() {
        vector<int> v(m_nums.begin(), m_nums.end());
        random_shuffle(v.begin(), v.end());
        return v;
    }
private:
    vector<int> m_nums;
};

总结

1， 北河的STL确实知道的很多，random_shuffle 惊艳了我

知识点

随机数

843. 猜猜这个单词@2018/06/26

北河


class Solution {
    vector<int> nums;
public:
    Solution(vector<int> nums) {
        this->nums = nums;
    }
    
    /** Resets the array to its original configuration and return it. */
    vector<int> reset() {
        return nums;
    }
    
    /** Returns a random shuffling of the array. */
    vector<int> shuffle() {
        vector<int> result(nums);
        for (int i = 0;i < result.size();i++) {
            int pos = rand()%(result.size()-i);
            swap(result[i+pos], result[i]);
        }
        return result;
    }
};

6. Z字形变换@2018/06/26

jzf

180. 连续出现的数字@2018/06/26

jzf

select distinct s1.Num as ConsecutiveNums from Logs as s1 ,Logs as s2,Logs as s3 
 where(( (s1.id= s2.id-1 and s1.id =s3.id-2)
        ||(s1.id =s2.id+1 and s1.id =s3.id-1)
        ||(s1.id =s2.id+2 and s1.id =s3.id+1))
       and(s1.Num =s2.Num and s1.Num =s3.Num))

7. 反转整数@2018/06/27

jzf

  • 第一版
class Solution {
public:
    int reverse(int x) {
        char str[16]={0};
        sprintf(str,"%d",x);
        string s =str;
        if(s[0]=='-')
        {
            std::reverse(s.begin()+1,s.end());
        }
        else
        {
            std::reverse(s.begin(),s.end());
        }
        long ret =atol(s.c_str());
        
        if(ret >INT_MAX||ret<INT_MIN)
        {
            return 0;
        }
        return  ret;
    }
};
  • 第二版 抄 ，我很喜欢这个解法
class Solution {
public:
    int reverse(int x) {
        long long res = 0;
        while (x != 0) {
            res = res * 10 + x % 10;
            x /= 10;
        }
        return (res < INT_MIN || res > INT_MAX) ? 0 : res;
    }
};
  • 第三版

小白

  • 第一版 出错
class Solution {
public:
int reverse(int x) {

	char str[12], *p = str + sprintf(str, "%d", x);

	std::reverse(str + (x < 0), p);

	int r = strtol(str, NULL, 10);

	return errno == ERANGE ? (errno = 0) : r;

}
};
  • 第二版
int reverse(int x) {
        auto str = to_string(x);
	std::reverse(str.begin(), str.end());
	try
	{
		return str.back() == '-' ? -std::stoi(str): std::stoi(str);
	}
	catch (...)
	{
		return 0;
	}
 }

总结

1，他这边用 strtol 和spritnf("%ld")出错 原因不知

知识点

1，这边atoi strtoi 遇见不合适的字符会跳过，和停止

147. 对链表进行插入排序@2018/06/27

北河 16ms

class Solution {
public:
    ListNode* insertionSortList(ListNode* head) {
        list<int> v;
        ListNode *p = head;
        while (p) { v.push_back(p->val); p = p->next;}
        v.sort();
        p = head;
        for (list<int>::iterator it = v.begin(); it != v.end(); ++it, p = p->next) {
            p->val = *it;
        }
        return head;
    }
};

jzf

class Solution {
public:
    ListNode* insertionSortList(ListNode* head) {
        ListNode* node =head;
        if(head ==NULL)
        {
            return head;
        }
        vector<ListNode*> vec;
        while(node != NULL)
        {
            vec.push_back(node);
            node =node->next;
        }
        sort(vec.begin(),vec.end(),[](ListNode*a,ListNode*b){return a->val <b->val;});
        for(auto iter =vec.begin();iter!= vec.end();iter++)
        {
            if(iter==vec.end()-1)
            {
                (*iter)->next =NULL;
                break;
            }
            (*iter)->next =*(iter+1);
            
        }
        return *vec.begin();
    }
};

puck


class Solution
{
  public:
    ListNode *insertionSortList(ListNode *head)
    {
        if (nullptr == head)
        {
            return nullptr;
        }
        
        ListNode *fast{head->next};
        ListNode *slow{head};
        ListNode *new_head{head};
        ListNode *tmp[2]{};

        while (nullptr != fast)
        {
            if (slow->val > fast->val)
            {
                // delete the node
                slow->next = fast->next;

                if (new_head->val > fast->val)
                {
                    fast->next = new_head;
                    new_head = fast;
                }
                else
                {
                    tmp[0] = new_head;
                    tmp[1] = new_head->next;
                    while (nullptr != tmp[1])
                    {
                        if (tmp[1]->val > fast->val)
                        {
                            fast->next = tmp[1];
                            tmp[0]->next = fast;
                            break;
                        }
                        tmp[0] = tmp[1];
                        tmp[1] = tmp[1]->next;
                    }
                }
            }
            
            slow = fast;
            fast = fast->next;
        }

        return new_head;
    }
};

828. 独特字符串@2018/06/28

北河

jzf

  • 第一版超时
class Solution {
public:
int uniqueLetterString(string S) {
	int sum = 0;
    vector<pair<int,string>> vec;
    for(size_t i =0;i<S.size();i++)
    {
        vector<pair<int,string>> vecTmp;
        vecTmp =vec;
        vec.clear();
        string str=S.substr(i,1);
        vec.push_back(make_pair(1,str));
        sum++;
        if(i ==0)
        {
            continue;
        }
        for(size_t j=0;j<i;j++)
        {
            int a =count(vecTmp[j].second.begin(),vecTmp[j].second.end(),S[i]);
            int b =0;
            if(a==0)
            {
                
                b =vecTmp[j].first+1;
                vec.push_back(make_pair(b,vecTmp[j].second+str));
            }
            else
            {
              if(a==1)
              {
                b =vecTmp[j].first-1;
               }
               else
               {
                   b =vecTmp[j].first;
               }
                vec.push_back(make_pair(b,vecTmp[j].second+str));
            }
            sum+=b;
        }
  
    }
	return sum;
}
};
  • 第二版超时

class Solution {
public:
int uniqueLetterString(string S) {
	int sum = 0;
    vector<pair<int,string>> vec;
    for(size_t i =0;i<S.size();i++)
    {
       
        string str=S.substr(i,1);
        vec.push_back(make_pair(1,str));
        sum++;
        if(i ==0)
        {
            continue;
        }
        for(size_t j=0;j<i;j++)
        {
            int a =count(vec[j].second.begin(),vec[j].second.end(),S[i]);
            if(a==0)
            {
                vec[j].second+=str;
                vec[j].first+=1;
                 sum+= vec[j].first;
            }
            else
            {
                if(a ==1)
                {
                    vec[j].first-=1;
                }
                vec[j].second+=str;
                sum+= vec[j].first;
            }
        }
  
    }
	return sum;
}
};

puck

#include <algorithm>
#include <unordered_map>

class Solution
{
public:
	int uniqueLetterString(string S)
	{
		std::size_t sz = S.size();
		int result{};

		auto ed = S.end();
		bool flg{true};
		for (auto iter{ S.begin() }; iter != ed; ++iter)
		{
			for (auto ite2{ 1 + iter }; flg; ++ite2)
			{
				if (ite2 == ed)
				{
					flg = false;
				}
				result += uniqueNum(string(iter, ite2));
			}
			flg = true;
		}

		return result;
	}

private:
	int uniqueNum(string s)
	{
		//std::unordered_map<char, int> map;
		//for (string::value_type ch : s)
		//{
		//	++map[ch];
		//}

		//int num{};
		//for (auto &pair : map)
		//{
		//	if (1 == pair.second)
		//	{
		//		++num;
		//	}
		//}

		//return num;
		
		short a[128]{};

		for (string::value_type ch : s)
		{
			++a[ch];
		}

		return std::count(std::begin(a), std::end(a), 1);
	}
};

693. 交替位二进制数@2018/06/29

jzf

  • 第一版 错误
class Solution {
public:
    bool hasAlternatingBits(int n) {
       // #define IS_TARGET(x)  ((x&0xaaaaaaaa)&&(x&0x55555555))
        int x =n;
        
        if( ((x&0xaaaaaaaa)==x)&&((x|0xaaaaaaaa)==0xaaaaaaaa))
        {
            int x =~n;
            if( ((x&0x55555555)==x)&&((x|0x55555555)==0x55555555))
           {
               return true;
           }
            else
            {
                return false;
            }
        }
        
        if( ((x&0x55555555)==x)&&((x|0x55555555)==0x55555555))
        {
            int x =~n; 
             if( ((x&0xaaaaaaaa)==x)&&((x|0xaaaaaaaa)==0xaaaaaaaa))
             {
                 return true;
             }
            else
            {
                return false;
            }
        }
        return false;
    }
};
  • 第二版 0s
class Solution {
public:
    bool hasAlternatingBits(int n) {
        
        int tmp=0;
        for(size_t i= n&1?0:1;i<31;)
        {
            tmp |=1<<i;
            if(tmp>n)
            {
                return false;
            }
            else if(tmp ==n)
            {
                return true;
            }
            i +=2;
        }
  
         return false;

    }
};

总结

1， 想通过按位与等方式解决问题的时候，对于异或同或之类的转换忘记了，结果绕了很大的弯子。数学还是要学一学的。

2，那个神奇的代码对提升速度真的很有效.

63. 不同路径 II@2018/06/30

北河 0ms


static auto XXX = []() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    return 0;
}();

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
        int m = obstacleGrid.size();
        int n = obstacleGrid[0].size();
        int v[m][n];
        memset(v, 0, m * n * sizeof(int));
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (obstacleGrid[i][j] != 0) {
                    continue;
                }
                if (i == 0 && j != 0) {
                    v[i][j] = v[i][j - 1];
                } else if (j == 0 && i != 0) {
                    v[i][j] = v[i - 1][j];
                } else if ( i != 0 && j != 0) {
                    v[i][j] += obstacleGrid[i - 1][j] > 0 ? 0 : v[i - 1][j];
                    v[i][j] += obstacleGrid[i][j - 1] > 0 ? 0 : v[i][j - 1];
                } else {
                   v[i][j] = 1;
                }
            }
        }
        return v[m - 1][n - 1];
    }
};

jzf

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
       int dp[101] ={0};
        for(int i =0;i<obstacleGrid.size();i++)
        {
            for(int j =0;j<obstacleGrid[i].size();j++)
            {
                if(i==0 &&j==0 )
                {
                    dp[j]=obstacleGrid[i][j]?0:1;
                }
                else if(j==0)
                {
                    dp[j]=obstacleGrid[i][j]?0:dp[j];
                }
                else
                {
                     dp[j]=obstacleGrid[i][j]?0:dp[j]+dp[j-1];
                }
            }
        }
        return dp[(*(obstacleGrid.end()-1)).size()-1];
    }
};

总结

1，先解决的718，我使用了暴力超时了，在摸索的过程中，我画了二叉树的图，和一个数组A和B为列表项的二维表。发现二维表可以很好的表示关系

2，根据二维表的规律，我学习了动态规划，但是还是多次失败了，其原因在于边界处理十分困难，如何处理二维表的边界上的计算公式很头疼。

718. 最长重复子数组@2018/07/01

jzf

  • 第一版 超时
class Solution {
public:
    int findLength(vector<int>& A, vector<int>& B) {
        int max=0;
      
     
        for(size_t i =0 ;i<A.size();i++)
        {  
           
            for(size_t j=0;j<B.size();j++)
            {
                 int tmp =0;
                size_t k =i;
                size_t x =j;
                while(k<A.size()&&x<B.size())
                {     
                    if(A[k]!=B[x]) 
                    {
                          break;
                    }
                    tmp++;
                    k++;
                    x++;
                 }
                if(tmp>max)
                {
                    max =tmp;
                }
            }
        }
        return max;
    }
};
  • 第二版 dp

北河

static auto XX = []() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    return 0;
}();

class Solution {
public:
    int findLength(vector<int>& A, vector<int>& B) {
        vector<int> dp(1001);
        int m = 0;
        for (int i = A.size() - 1; i >= 0; --i) {
            for (size_t j = 0; j < B.size(); ++j) {
                if (A[i] == B[j]) {
                    dp[j] = dp[j + 1] + 1;
                    if (dp[j] > m) {
                        m = dp[j];
                    }
                } else {
                    dp[j] = 0;
                }
            }
        }
        return m;
    }
};

406. 根据身高重建队列@2018/07/02 未备份

455. 分发饼干@2018/07/03

jzf

class Solution {
public:
    int findContentChildren(vector<int>& g, vector<int>& s) {
        sort(g.begin(),g.end());
        sort(s.begin(),s.end());
        
        int j =0;
        int i=0;
        int t =0;
        for(;i<std::min(g.size(),s.size());i++)
        {
            for(;j<s.size();)
            {
                if(g[i] <=s[j])
                {
                    t++;   
                    j++;
                    break;
                }
                j++;
            }
        }
        return t;
    }
};

北河

738.单调递增的数字@2018/07/04

北河

jzf


class Solution {
public:
    int turn2Int(char *p)
    {
        int ret=0;
        while(*p !=0)
        {
            ret= ret*10+(*p-'0');
            p++;
        }
        return ret;
    }
    void tostr(int tmp,char* p)
    {
        int s =1;
        int q =tmp;
        for(;q>10;q/=10)
        {
            s*=10;
        }
        for(;s>0;s/=10)
        {
            *(p++)=tmp/s+'0';
            tmp=tmp%s;
        }
    }
    int monotoneIncreasingDigits(int N) {
       // tostring(N);
        char str[11]={0};
        tostr(N,str);
        
        int t=0;
        for(size_t i=0;i<11 &&str[i]!=0;i++)
        {
            if(str[i] > str[i+1])
            {
                t =i;
                break;
            }
        }
        if(t==0)
        {
            str[0]-=1;
            for(int i= t+1;i<11 &&str[i]!=0;i++)
            {
                str[i] ='9';
            }
             return turn2Int(str+t);
        }
        else if(str[t+1]==0)
        {
            return  N;cout<<12;
        }
        else
        {
            int tmp =t;
            while(t>0)
            {
                    if(str[t]>str[t-1])
                    {
                         str[t]-=1;
                        for(int i= t+1;i<11 &&str[i]!=0;i++)
                        {
                            str[i] ='9';
                        }
                        return turn2Int(str);
                    }
                t--;
            }
            
            if(t==0)
            {
               
                  str[0]-=1;
                for(int i= t+1;i<11 &&str[i]!=0;i++)
                 {
                    str[i] ='9';
                }
                return turn2Int(str+t);
            }
           
        }
        return turn2Int(str+t);
    }
};

313. 超级丑数@2018/07/05 无人做

20. 有效的括号@2018/07/06

羽柔子

var isValid = function (s) {
    var l = new Array();
    return s.split("").every(e => {
        switch (e) {
            case '(': return l.push(1);
            case ')': return l.pop() == 1;
            case '{': return l.push(2);
            case '}': return l.pop() == 2;
            case '[': return l.push(3);
            case ']': return l.pop() == 3;
        };
    }) ? !l.length : false;
};

jzf

  • 第一版错误
class Solution {
public:
    bool isClosed(char c, char dst)
 {
	 switch (c)
	 {
	 case '[':
		 return dst == ']';
	 case '{':
		 return dst == '}';
	 case '(':
		 return dst == ')';
	 default:
		 return false;
	 }
 }
 bool isOpen(char c)
 {
	 switch (c)
	 {
	 case '[':

	 case '{':

	 case '(':
		 return true;
	 default:
		 return false;
	 }
 }
 bool check(size_t &i, string &s)
 {
	 size_t j = i + 1;
     bool ret=false;
	 for (; j<s.size(); j++)
	 {
		 if (isClosed(s[i], s[j]) && ((j - i) & 1) == 1)
		 {
             ret =true;
			 for (size_t k = (j - i-1)/2; k>0; k--)
			 {
                 
				 if (!isClosed(s[i + k], s[j - k]))
				 {
					 return false;
				 }
			 }
             break;
		 }
	 }
	 i = j;
	 return ret;
 }
 bool isValid(string s) {
	 // '[''{''('')''}'']'
	 for (size_t i = 0; i<s.size(); i++)
	 {
		 if (isOpen(s[i]) && (i & 1) == 0)
		 {
			 if (!check(i, s))
			 {
				 return false;
			 }
		 }
		 else
		 {
			 return false;
		 }
	 }
	 return true;
 }
};
  • 第二版
class Solution {
public:
    bool isClosed(char c, char dst)
 {
	 switch (c)
	 {
	 case '[':
		 return dst == ']';
	 case '{':
		 return dst == '}';
	 case '(':
		 return dst == ')';
	 default:
		 return false;
	 }
 }
 bool isOpen(char c)
 {
	 switch (c)
	 {
	 case '[':

	 case '{':

	 case '(':
		 return true;
	 default:
		 return false;
	 }
 }

 bool isValid(string s) {
	 // '[''{''('')''}'']'
     bool ret =false; 
     int tmp =0; 
    while(s.size()!=tmp)
    {
        tmp =s.size();
        for(int i=0;i<tmp-1;i++)
        {
            if(isClosed(s[i],s[i+1]))
            {
                s.erase(i,2);
                break;
            }
        }
    }
     return tmp==0?true:false;
 }
};

三浪

typedef struct tagStack
{
	char * p;
	int iPos;
}Stack;

void BulidStack(Stack *pStack, int size)
{
	pStack->p = (char *)malloc(size * sizeof(char));
	memset(pStack->p, 0, size);
	pStack->iPos= 0;
}

void DestroyStack(Stack *pStack)
{
	free(pStack->p);
}

int Push(Stack *pStack, char ch)
{
	if (pStack->iPos >= 0)
	{
		*(pStack->p + pStack->iPos) = ch;
		++(pStack->iPos);
		return 0;
	}
	return -1;
}

int Pop(Stack *pStack)
{
	--(pStack->iPos);
	if (pStack->iPos >= 0)
	{
		*(pStack->p + pStack->iPos) = '\0';
		return 0;
	}
	return -1;
}
char GetPairChar(char ch)
{
	switch (ch)
	{
		case ')':
			return '(';
		case '}':
			return '{';

		case ']':
			return '[';
		default:
			return '\0';
	}
}

bool IsPush(char ch)
{
	if (ch == '(' || ch == '{' || ch == '[')
	{
		return true;
	}
	return false;
}

bool isValid(char *s)
{
	if ('\0' == *s)
	{
		return true;
	}
	int n = 0;
	char chPre = *s;
	if (!IsPush(chPre))
	{
		return false;
	}

	Stack myStack;
	BulidStack(&myStack, strlen(s) + 1);
	while (*s != '\0')
	{
		char chCur = *s;
		if (IsPush(chCur))
		{
			if (0 != Push(&myStack, chCur))
			{
				return false;
			}
		}
		else
		{
			if ((myStack.iPos >= 1) && 
			(myStack.p[myStack.iPos - 1] == GetPairChar(chCur)))
			{
				if (0 != Pop(&myStack))	
				{
					return false;
				}
			}
			else
			{

				if (0 != Push(&myStack, chCur))
				{
					return false;
				}
			}
		}
		chPre = *s;
		++s;
	}
	if (0 == myStack.iPos)
	{
		DestroyStack(&myStack);
		return true;
	}
	else
	{
		DestroyStack(&myStack);
		return false;
	}

}

总结

1，我在做这道题的时候，出现了许多次的分析失误，疏忽了一些情形。 2，发现 s.size()的返回类型是size_t ，遇到了一个比较悲惨的越界错误。 size_t是unsigned int类型的 当 for(int i=0; i<tmp -1;i++)中的tmp = 0时，就出现了大bug;