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.
思路：使用两个指针分别指向当前窗口的起始位置，使用哈希表charT存储t中字符出现的次数，使用哈希表find存储当前窗口中在t中出现的字符出现的次数（256个字符，使用数组代替哈希表即可），使用findCount记录找到在t中出现字符的个数（注意t中的字符可能会出现重复）。具体步骤：
(1)初始化charT
(2)遍历s，若charT[s[i]]非0，即s[i]出现在t中，find[s[i]]++,若find[s[i]]< charT[s[i]],findCount++
(3)如果findCount==len(t)，表示找到一个符合的窗口，此时移动start，如果charT[s[start]]为0，则start++;若find[s[start]] > charT[s[start]]，则find[s[start]] –，同时start++;否则停止移动start，此时s[start,i]即为满足条件的窗口。
(4)重复(2)(3)，直至找到满足条件的最小窗口

class Solution {public:    string minWindow(string s, string t) {        int lent = t.length();        int lens = s.length();        if(lent == 0 || lens == 0)            return "";        int charT[256] = {0};//t中的字符出现的次数        int find[256] = {0};//窗口中在t中出现字符出现的次数        int findCount = 0, start = 0, i;        string win = "";        for(i = 0; i < lent; ++i){            charT[t[i]]++;        }        i = 0;        while(i < lens){            if(charT[s[i]]){                find[s[i]] ++;                if(find[s[i]] <= charT[s[i]])                    findCount ++;                if(findCount == lent){//找到一个窗口                    while(1){                        if(charT[s[start]] == 0)                            start ++;                        else if(find[s[start]] > charT[s[start]])                            find[s[start++]]--;                        else                            break;                    }                    int a = i - start + 1;                    if(win == "")                        win = s.substr(start, a);                    else                        win = win.length() > a ? s.substr(start, a) : win;                    find[s[start]]--;                    start ++;                    findCount --;                }            }            i++;        }        return win;    }};