[LeetCode] 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).

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.

 

两个指针，start，end。end往后面搜。存在满足要求的后，尽量压缩start。

 

class Solution {
public:
    string minWindow(string S, string T) {
        int c[256] = {0};
        bool cb[256] = {0};
        int cnt = T.size();
        for (char i: T) c[(int)i]++, cb[(int)i] = true;
        
        int start = 0, end = 0;
        int ls = S.size();
        int minlen = 10000;
        string minstr = "";
        while (end < ls && start < ls) {
            if (cb[S[end]]) {
                c[S[end]]--;
                if (c[S[end]] >= 0) cnt--;
                while (cnt == 0) {
                    if (cb[S[start]] == false) start++;
                    else if (c[S[start]] < 0) {
                        c[S[start]]++;
                        start++;
                    } else {
                        break;
                    }
                }
                if (cnt == 0) {
                    if (end - start + 1 < minlen) {
                        minlen = end - start + 1;
                        minstr = S.substr(start, minlen);
                    }
                    c[S[start]]++;
                    start++;
                    cnt++;
                }
            }
            end++;
        }
        return minstr;
    }
};