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.

思路：因为O（n）的时间复杂度，所以考虑用双指针。

class Solution {public:    string minWindow(string S, string T) {        int tmplate[128] = {0};        int find_chars[128] = {0};        int t_count = T.size();                for(int i = 0; i < t_count; i++) {            tmplate[T[i]]++;        }                int min_length = S.size() + 1;        int min_start = -1;        int count = 0;        int start = 0; // start相当于前指针        for(int i = 0; i < S.size(); i++) { //i是后指针            if(tmplate[S[i]] == 0){                continue;            }                        find_chars[S[i]]++;            if(find_chars[S[i]] <= tmplate[S[i]]) {                count++;            }                        while(count == t_count) { //当找到一次完整的tmplate后处理                while(tmplate[S[start]] == 0) {                    start++;                }                                int tmp_length = i - start + 1; //得到当前最小的subsr                if(tmp_length < min_length) {                    min_length = tmp_length;                    min_start = start;                }                                find_chars[S[start]]--; //移动前指针的三个步骤，首先对应的这个字符的统计个数减去1                if(find_chars[S[start]] < tmplate[S[start]]) {  //去除前指针对应的字符后，对组合成template的影响，如果变少，count要减                    count--;                }                start++; //移动前指针            }        }                return min_start == -1 ? "": S.substr(min_start, min_length);    }};

Ref: http://blog.csdn.net/ymhhym/article/details/19432339