Leetcode231: 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 empty string "".

If there are multiple such windows, you are guaranteed that there will always be only one unique minimum window in S.

双指针思想，尾指针不断往后扫，当扫到有一个窗口包含了所有T的字符，然后再收缩头指针，直到不能再收缩为止。最后记录所有可能的情况中窗口最小的

class Solution {private:    int count1[256];    int count2[256];public:    string minWindow(string S, string T) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        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]]++;        }                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--;            }            //找到了包含所有字符的end位置            if (count == 0)            {//从起点向后压缩，直到不能压缩为止                while(true)                {                    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 "";                string ret(S, minStart, minSize);                return ret;            }};