leetcode: Minimum Window Substring

最新推荐文章于 2025-02-20 13:41:31 发布

原创最新推荐文章于 2025-02-20 13:41:31 发布 · 313 阅读

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本文提供了一种复杂度为O(n)的方法来查找字符串中包含特定字符集的最短子串。通过使用哈希映射记录字符出现情况，实现高效查找并确定最小覆盖子串。

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同时请参阅：http://blog.youkuaiyun.com/martin_liang/article/details/8592894

相同的问题，解法稍微有一点不一样

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.

class Solution {
public:
    string minWindow(string S, string T) {
        
        map<char, int> matchingMap;
        
        for (int i=0; i<T.size(); i++)
        {
            if (matchingMap.find(T.at(i)) == matchingMap.end())
            {
                matchingMap[T.at(i)] = 1;
            }
            else
            {
                matchingMap[T.at(i)]++;
            }
        }
        
        map<char, int> loopingMap;
        int count = 0;
        int startPos = 0;
        int minLen = S.size();
        string retString;
        for (int i=0; i<S.size(); i++)
        {
            if (matchingMap.find(S.at(i)) != matchingMap.end())
            {
                if (loopingMap.find(S.at(i)) == loopingMap.end())
                {
                    loopingMap[S.at(i)] = 1;
                }
                else
                {
                    loopingMap[S.at(i)]++;
                }
                
                if (loopingMap[S.at(i)] <= matchingMap[S.at(i)])
                    count++; // example  T:ABC   S: AAABC, 当遇到第二和第三个A时，count都不会加一，最终要进行match的时候，count必定为3                    
                if (count == T.size())
                {
                    while (true) // 注意这里的循环
                    {
                        if (matchingMap.find(S.at(startPos)) == matchingMap.end())
                        {
                            startPos++;
                            continue;
                        }
                
                        if (loopingMap[S.at(startPos)] > matchingMap[S.at(startPos)])
                        {
                            loopingMap[S.at(startPos)]--;
                            startPos++;
                        }
                        else
                        {
                            break; 
                        }
                    }
                
                    if (i-startPos+1 <= minLen)
                    {
                        minLen = i-startPos+1;
                        retString = S.substr(startPos, minLen);
                    }
                
                    count--;
                    loopingMap[S.at(startPos)]--;
                    startPos++;
                }
            }
        }
        return retString;
    }
};