76. Minimum Window Substring

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.

 

AC code:

class Solution {
public:
    string minWindow(string s, string t) {
        map<char, int> m;
        for (int i = 0; i < t.length(); ++i) 
            m[t[i]]++;
        int total = t.length();
        int from = 0;
        int minn = INT_MAX;
        for (int i = 0, j = 0; i < s.length(); ++i) {
            if (m[s[i]]-- > 0) total--;
            while (total == 0) {
                if (i-j+1 < minn) {
                    minn = i-j+1;
                    from = j;
                }
                if (++m[s[j++]] > 0) total++;    
            }
        }
        return (minn == INT_MAX) ? "" : s.substr(from, minn);
    }
};

Runtime: 20 ms, faster than 45.51% of C++ online submissions for Minimum Window Substring.

 

c++string.find（）函数用法整理

C++ string substr()