LeetCode-76. Minimum Window Substringhttps://leetcode.com/problems/minimum-window-substring/

题目描述

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.

A substring is a contiguous sequence of characters within the string.

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.

解题思路

【C++】

class Solution {
public:
    string minWindow(string s, string t) {
        vector<int> ch(256, 0);
        for(auto c : t) ch[c]++;
        int b=0, e=0, cnt=t.size(), minStart=0, minL=INT_MAX;
        while (e<s.size()) {
            if(ch[s[e++]]-- > 0) cnt--;
            while(cnt == 0) {
                if(e-b < minL) {minL=e-b; minStart=b;}
                if(ch[s[b++]]++ == 0) cnt++;
            }
        }
        return minL == INT_MAX ? "" : s.substr(minStart, minL);
    }
}

【Java】

class Solution {
    public String minWindow(String s, String t) {
        int[] ch = new int[256];
        for(char c : t.toCharArray()) ch[c]++;
        int b=0, e=0, cnt=t.length(), minStart=0, minL=Integer.MAX_VALUE;
        while (e<s.length()) {
            if(ch[s.charAt(e++)]-- > 0) cnt--;
            while(cnt == 0) {
                if(e-b < minL) {minL=e-b; minStart=b;}
                if(ch[s.charAt(b++)]++ == 0) cnt++;
            }
        }
        return minL == Integer.MAX_VALUE ? "" : s.substring(minStart, minStart + minL);
    }
}

参考文献

【1】leetcode刷题2———子串系列_taifyang的博客-CSDN博客