面试经典150题——Day33

经典150题——Day33"/>

面试经典150题——Day33

文章目录

• • 一、题目
  • 二、题解

一、题目

76. Minimum Window Substring

Given two strings s and t of lengths m and n respectively, return the minimum window
substring
of s such that every character in t (including duplicates) is included in the window. If there is no such substring, return the empty string “”.

The testcases will be generated such that the answer is unique.

Example 1:

Input: s = “ADOBECODEBANC”, t = “ABC”
Output: “BANC”
Explanation: The minimum window substring “BANC” includes ‘A’, ‘B’, and ‘C’ from string t.
Example 2:

Input: s = “a”, t = “a”
Output: “a”
Explanation: The entire string s is the minimum window.
Example 3:

Input: s = “a”, t = “aa”
Output: “”
Explanation: Both 'a’s from t must be included in the window.
Since the largest window of s only has one ‘a’, return empty string.

Constraints:

m == s.length
n == t.length
1 <= m, n <= 105
s and t consist of uppercase and lowercase English letters.

Follow up: Could you find an algorithm that runs in O(m + n) time?

题目来源： leetcode

二、题解

滑动窗口，两个哈希表

class Solution {
public:string minWindow(string s, string t) {string res = s;int n1 = s.length(),n2 = t.length();if(n1 < n2) return "";unordered_map<char,int> tmap;unordered_map<char,int> window;for(int i = 0;i < n2;i++) tmap[t[i]]++;int i = 0,j = 0;int begin = 0;int resLen = s.length();int cnt = 0;bool exist = false;while(j < n1){//如果未满，则扩大jif(tmap[s[j]] == 0){j++;continue;}if(window[s[j]] < tmap[s[j]]) cnt++;window[s[j]]++;j++;cout << 111 << endl;while(cnt == n2){if(j - i <= resLen){begin = i;resLen = j - i;exist = true;}//缩小iif(tmap[s[i]] == 0){i++;continue;}if(window[s[i]] == tmap[s[i]]){cnt--;}window[s[i]]--;i++;}}res = s.substr(begin,resLen);return exist ? res : "";}
};