[LeetCode] Minimum Window Substring

最新推荐文章于 2025-09-11 19:19:55 发布

angelof7

最新推荐文章于 2025-09-11 19:19:55 发布

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分类专栏： LeetCode 文章标签： leetcode substring

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本文详细介绍了如何使用双指针技术在复杂度为O(n)的情况下，找到给定字符串中包含所有特定子串字符的最短子串。通过实例演示和代码实现，深入探讨了解题思路和细节。

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 emtpy string "".

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

利用双指针思想，尾指针不断往后扫，当扫到有一个窗口包含了所有T的字符，然后再收缩头指针，直到不能再收缩为止。

然后再向后扩展，再收缩头指针。。。

最后记录所有可能的情况中窗口最小的

public class Solution {
    public String minWindow(String S, String T) {
        int[] count1 = new int[256];
        int[] count2 = new int[256];
        
        for(int i = 0; i < 256; i++){
            count1[i] = count2[i] = 0;
        }
        int count = T.length();
        for(int i = 0; i < count; i++){
            count1[T.charAt(i)]++;
            count2[T.charAt(i)]++;
        }
        
        int start = 0;
        int minSize = S.length() + 1;
        int minStart = 0;
        
        for(int i = 0; i < S.length(); i++){
            if(count2[S.charAt(i)] > 0){
                count1[S.charAt(i)]--;
                if(count1[S.charAt(i)] >= 0)
                    count--;
            }
            
            if(count == 0){
                while(true){
                    if(count2[S.charAt(start)] > 0){
                        if(count1[S.charAt(start)] < 0)
                            count1[S.charAt(start)]++;
                        else
                            break;
                    }
                    start++;
                }
                
                if(minSize > i - start + 1){
                    minSize = i -start + 1;
                    minStart = start;
                }
            }
        }
        
        if(minSize == S.length() + 1)
            return "";
        return S.substring(minStart, minStart + minSize);
    }
}