做企业平台的网站有哪些短视频seo是什么

Problem: 76. 最小覆盖子串

题目描述

思路

1.定义两个map集合need和window（以字符作为键，对应字符出现的个数作为值），将子串t存入need中；
2.定义左右指针left、right均指向0（形成窗口），定义int类型变量len记录最小窗口长度,valid记录当前窗口否存最短子串中的字符个数；
3.向右扩大窗口，遍历到到的字符c如果在need中时，window[c]++，同时如果window[c] == need[c]，则valid++；
4.如果valid == need.size()，则表示可以开始收缩窗口，并更新最小窗口禅读（如果移除的字符在need中，同时window[d] == need[d],则valid–，window[d]–）;

复杂度

时间复杂度:

$O(n)$;其中$n$为字符串$s$的长度

空间复杂度:

$O(n)$

Code

class Solution {
public:/*** Two pointer** @param s Given string* @param t Given string* @return string*/string minWindow(string s, string t) {unordered_map<char, int> need;unordered_map<char, int> window;for (char c: t) {need[c]++;}int left = 0;int right = 0;int valid = 0;// Records the starting index and length of the minimum overlay substringint start = 0;int len = INT_MAX;while (right < s.size()) {//c is the character moved into the windowchar c = s[right];// Move the window rightright++;// Perform some column updates to the data in the windowif (need.count(c)) {window[c]++;if (window[c] == need[c]) {valid++;}}// Determine whether to shrink the left windowwhile (valid == need.size()) {// Update the minimum overlay substringif (right - left < len) {start = left;len = right - left;}//d is the character to be moved out of the windowchar d = s[left];// Move the window leftleft++;// Perform some column updates to the data in the windowif (need.count(d)) {if (window[d] == need[d]) {valid--;}window[d]--;}}}// Returns the minimum overlay substringreturn len == INT_MAX ? "" : s.substr(start, len);}
};