题目：

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.

代码：

方法1——用map：

class Solution {
public:
    string minWindow(string s, string t) {
        string res = "";
        unordered_map<char, int> letterCnt;
        int left = 0, cnt = 0, minLen = INT_MAX;
        for (char c : t) ++letterCnt[c];
        for (int i = 0; i < s.size(); ++i) {
            if (--letterCnt[s[i]] >= 0) ++cnt;
            while (cnt == t.size()) {
                if (minLen > i - left + 1) {
                    minLen = i - left + 1;
                    res = s.substr(left, minLen);
                }
                if (++letterCnt[s[left]] > 0) --cnt;
                ++left;
            }
        }
        return res;
    }
};

方法二——用vector:

class Solution {
public:
    string minWindow(string s, string t) {
        string res = "";
        vector<int> letterCnt(128, 0);
        int left = 0, cnt = 0, minLen = INT_MAX;
        for (char c : t) ++letterCnt[c];
        for (int i = 0; i < s.size(); ++i) {
            if (--letterCnt[s[i]] >= 0) ++cnt;
            while (cnt == t.size()) {
                if (minLen > i - left + 1) {
                    minLen = i - left + 1;
                    res = s.substr(left, minLen);
                }
                if (++letterCnt[s[left]] > 0) --cnt;
                ++left;
            }
        }
        return res;
    }
};