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.
题意：找到S串中包含T的最短的子串，可以不按顺序
思路：还是双指针发处理，尾指针先找到已经包含了T串的位置，然后头指针再不停的收缩，标记的方式就是记录字符出现的次数

class Solution {
public:
    int count1[256];
    int count2[256];

    string minWindow(string S, string T) {
        if (T.size() == 0 || S.size() == 0)
            return "";
            
        memset(count1, 0, sizeof(count1));
        memset(count2, 0, sizeof(count2));
        
        for(int i = 0; i < T.size(); i++) {
            count1[T[i]]++;
            count2[T[i]] = 1;
        }
        
        int count = T.size();
        
        int start = 0;
        int minSize = INT_MAX;
        int minStart;
        for(int end = 0; end < S.size(); end++) {
            if (count2[S[end]] > 0) {
                count1[S[end]]--;
                if (count1[S[end]] >= 0)
                    count--;
            }
            
            if (count == 0) {
                while(1) {
                    if (count2[S[start]] > 0) {
                        if (count1[S[start]] < 0)
                            count1[S[start]]++;
                        else break;
                    }
                    start++;
                }
            
                if (minSize > end - start + 1) {
                    minSize = end - start + 1;
                    minStart = start;
                }
            }
        }
        
        if (minSize == INT_MAX) return "";
        return S.substr(minStart, minSize);
    }
};