网站建设的广告投入,建立的英语,硬件开发平台有哪些,网站建设网站设计Problem: 76. 最小覆盖子串 文章目录 题目描述思路复杂度Code 题目描述 思路 1.定义两个map集合need和window#xff08;以字符作为键#xff0c;对应字符出现的个数作为值#xff09;#xff0c;将子串t存入need中#xff1b; 2.定义左右指针left、right均指向0#xff… Problem: 76. 最小覆盖子串 文章目录 题目描述思路复杂度Code 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 ) O(n) O(n);其中 n n n为字符串 s s s的长度 空间复杂度: O ( n ) O(n) 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_mapchar, int need;unordered_mapchar, 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);}
};
