[LeetCode] Minimum Window Substring

最新推荐文章于 2022-12-06 16:35:15 发布

原创最新推荐文章于 2022-12-06 16:35:15 发布 · 438 阅读

0 ·

CC 4.0 BY-SA版权

leetcode 专栏收录该内容

89 篇文章

订阅专栏

本文介绍了一种在字符串S中寻找包含字符串T所有字符的最短子串的算法，复杂度为O(n)。通过使用两个固定大小的数组来跟踪目标字符及已匹配字符的数量，实现了高效查找。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Given a string S and a string T, find the minimum window in S which will contain all the characters in T in complexity O(n).

For example,
S = "ADOBECODEBANC"
T = "ABC"

Minimum window is "BANC".

Note:
If there is no such window in S that covers all characters in T, return the empty string "".

If there are multiple such windows, you are guaranteed that there will always be only one unique minimum window in S.

滑动子窗口(为了让代码精简，用int[128]，不要用HashMap)：

public class Solution {
    public String minWindow(String s, String t) {
		int[] map=new int[128];
		int[] map2=new int[128];
		for(int i=0;i<t.length();i++) map[t.charAt(i)]++;
		int n=0;
		int reStart=0,reEnd=Integer.MAX_VALUE;
		int start=0;
		for(int i=0;i<s.length();i++){
			char ch=s.charAt(i);
			if(map[ch]>0){
			    if(map2[ch]<map[ch]) n++;
				map2[ch]++;
			}
			if(n>=t.length()){
				int k=start;
				for(;k<=i;k++){
					char c=s.charAt(k);
					if(map[c]==0) continue;
					if(map2[c]>map[c]){
						map2[c]--;
					}else break;
				}
				start=k;
				if(reEnd-reStart>i-start){
					reEnd=i;
					reStart=start;
				}
			}
		}
		if(reEnd==Integer.MAX_VALUE) return "";
		else return s.substring(reStart,reEnd+1);
	}
}