leetcode || 76、Minimum Window Substring

最新推荐文章于 2025-12-21 17:04:13 发布

原创最新推荐文章于 2025-12-21 17:04:13 发布 · 401 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode #双指针 #substring #string #算法

LeetCode 专栏收录该内容

131 篇文章

订阅专栏

本文介绍了一种在给定字符串中寻找包含另一字符串所有字符的最短子串的算法，采用双指针技巧实现复杂度为O(n)。通过不断扩展尾指针直至覆盖目标字符串，再收缩头指针来确定最短窗口。

problem：

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.

Hide Tags

Hash Table Two Pointers String

题意：在一个字符串中寻找一个包含目标字符串的最短的子字符串

thinking：

（1）思路很简单，双指针思想，尾指针不断往后搜索，直到包含字符串T，收缩头指针。记录尾头指针间距最小的位置，输出

code：

class Solution {
private:
    int count1[256];
    int count2[256];
public:
    string minWindow(string S, string T) {
        if (T.size() == 0 || S.size() == 0)
            return "";
            
        memset(count1, 0, sizeof(count1));
        memset(count2, 0, sizeof(count2));
        
        for(int i = 0; i < T.size(); i++)
        {
            count1[T[i]]++;
            count2[T[i]]++;
        }
        
        int count = T.size();
        
        int start = 0;
        int minSize = INT_MAX;
        int minStart;
        for(int end = 0; end < S.size(); end++)
        {
            if (count2[S[end]] > 0)
            {
                count1[S[end]]--;
                if (count1[S[end]] >= 0)
                    count--;
            }
            
            if (count == 0)
            {
                while(true)
                {
                    if (count2[S[start]] > 0)
                    {
                        if (count1[S[start]] < 0)
                            count1[S[start]]++;
                        else
                            break;
                    }
                    start++;
                }
            
                if (minSize > end - start + 1)
                {
                    minSize = end - start + 1;
                    minStart = start;
                }
            }
        }
        
        if (minSize == INT_MAX)
            return "";
        
        string ret(S, minStart, minSize);
        
        return ret;        
    }
};