51 KiB
51 KiB
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;