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).

Example:

Input: S = "ADOBECODEBANC", T = "ABC"
Output: "BANC"

Note:

• If there is no such window in S that covers all characters in T, return the empty string​​""​​.
• If there is such window, you are guaranteed that there will always be only one unique minimum window in S.

题解：

滑动窗口，首先计算出t中所有字符出现次数，用map存储，然后遍历s，当遍历到符合要求时，保存当前left，rigth和长度，即t中所有字符都出现了，左边开始收缩窗口，收缩到t中所有字符不出现，即不满足要求，然后再往下遍历s寻找下一个符合要求的串。

class Solution {
public:
    string minWindow(string s, string t) {
        int n = s.length(), num = t.length();
        unordered_map<char, int> mp;
        for (int i = 0; i < num; i++) {
            mp[t[i]]++;
        }
        int len = n + 1, appear = 0, left = 0, begin = 0;
        for (int i = 0; i < n; i++) {
            if (mp.find(s[i]) != mp.end()) {
                mp[s[i]]--;
                if (mp[s[i]] >= 0) {
                    appear++;
                }
            }
            while (appear == num) {
                if (i - left + 1 < len) {
                    begin = left;
                    len = i - left + 1;
                }
                if (mp.find(s[left]) != mp.end()) {
                    mp[s[left]]++;
                    if (mp[s[left]] > 0) {
                        appear--;
                    }
                }
                left++;
            }
        }
        if (len == n + 1) {
            return "";
        }
        return s.substr(begin, len);
    }
};