14 KiB
14 KiB
create by jzf@2018/06/20
[TOC]
作业
[TOC]
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
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
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
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;
}
};
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;
}
};
北河
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);