[Leetcode] Minimum Window Substring-优快云博客

本文链接：https://blog.youkuaiyun.com/liuwenyirachel/article/details/8583999

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)。通过使用双指针技术和哈希表记录字符出现次数，确保了高效查找。文章提供了详细的实现代码。

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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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function        
        int sizeS = S.size();
        int sizeT = T.size();
        int length = INT_MAX;
        
        int position = 0;
        int countChar = 0;
        
        vector<int> vectorMap(256,0);
        
        for(int i=0;i<sizeT;i++)
        {
            vectorMap[T[i]]++; //构造字符哈希
        }
        
        for(int i=0;i<256;i++)
        {
            if(vectorMap[i])
            {
                countChar++; //T中共有几个不同的字符
            }            
        }
        
        int j = 0; //双指针i与j，j指向头，i指向尾。
        
        for(int i=0;i<sizeS;i++)
        {
            vectorMap[S[i]]--;
            if(vectorMap[S[i]]==0) 
            {
                countChar--;
            }            
            while(vectorMap[S[j]]<0) //多余的字符
            {
                vectorMap[S[j]]++;
                j++;
            }
            if(countChar==0&&i-j+1<length)
            {
                length = i-j+1;
                position = j;
            }    
        }        
        return (length==INT_MAX)?"":S.substr(position,length);         
    }
};