Minimum Window Substring [leetcode]-优快云博客

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)。通过双指针技术和字符计数方法实现了高效查找，并给出了具体实现代码。

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.

string minWindow(string S, string T) {
        string res = "";
        int tCharNum[256] = {0};
        int sCharNum[256] = {0};
        int l = 0,r = 0;
        
        for (int i = 0; i < T.size(); i++)
            tCharNum[T[i]] ++;
        
        while (r < S.size())
        {
            if (tCharNum[S[r]])
            {
                sCharNum[S[r]]++;
                while (check(sCharNum, tCharNum))
                {
                    if (res == "" || res.size() > (r - l + 1)) res = S.substr(l, r - l + 1);
                    if (sCharNum[S[l]])     sCharNum[S[l]]--;
                    do { l++; } while (l < S.size() && tCharNum[S[l]] == 0);
                }
            }
            r++;
        }
        return res;
    }
    
    bool check(int * t1, int * t2)
    {
        for (int i = 0; i < 256; i++)
            if (t1[i] < t2[i]) return false;
        return true;
    }