LeetCode: Minimum Window Substring

最新推荐文章于 2021-02-25 13:50:45 发布

原创最新推荐文章于 2021-02-25 13:50:45 发布 · 1.9k 阅读

0 ·

CC 4.0 BY-SA版权

算法同时被 2 个专栏收录

109 篇文章

订阅专栏

C++

108 篇文章

订阅专栏

本文介绍了一种使用字符哈希结合双指针技术，在O(n)复杂度内找到字符串S中包含字符串T所有字符的最小子串的方法。通过实例演示了如何通过计数和滑动窗口来高效解决此问题。

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 emtpy string "".

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

字符哈希+双指针。

class Solution {
public:
    string minWindow(string S, string T) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int nT = T.size();
        int nS = S.size();
        
        int needToFind[256] = {0};
        for (int i = 0; i < nT; ++i)
            ++needToFind[T[i]];
        int hasFound[256] = {0};
        int minBegin;
        int minEnd;
        int minWindow = nS + 1;
        int count = 0;
        
        char ch;
        for (int begin = 0, end = 0; end < nS; ++end)
        {
            if (needToFind[S[end]] == 0)
                continue;
            ch = S[end];
            ++hasFound[ch];
            if (hasFound[ch] <= needToFind[ch])
                ++count;
                
            if (count == nT)
            {
                while (needToFind[S[begin]] == 0
                || hasFound[S[begin]] > needToFind[S[begin]])
                {
                    if (hasFound[S[begin]] > needToFind[S[begin]])
                        --hasFound[S[begin]];
                    ++begin;
                }
                
                int length = end - begin + 1;
                if (length < minWindow)  
                {
                    minBegin = begin;
                    minEnd = end;
                    minWindow = length;
                }
            }
        }
        
        return minWindow <= nS ? S.substr(minBegin, minEnd-minBegin+1) : "";
    }
};