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：涉及到2点： 1. Hash, 保证查找为 O(1). 2. S 中设置两指针，根据长度确定右边指针位置；根据若去掉该字符，则该字符在 window 中出现次数将小于在 T 中出现的次数确定左边指针位置。
class Solution {
public:
string minWindow(string S, string T) {
if(T == "" || S == "") return "";
string s;
int ch[2][‘z‘+1] = {0};
for(int i = 0; i < T.size(); ++i) ++ch[0][T[i]];
int first = 0, cnt = 0;
for(int second = 0; second < S.size(); ++second) {
if(ch[0][S[second]]) {
if(ch[0][S[second]] > ch[1][S[second]])
++cnt;
++ch[1][S[second]];
}
if(cnt == T.size()) {
while(ch[0][S[first]] == 0 || ch[1][S[first]] > ch[0][S[first]]) {
if(ch[1][S[first]] > ch[0][S[first]])
ch[1][S[first]]--;
++first;
}
string tem = S.substr(first, second-first+1);
if(s == "" || s.size() > tem.size()) s = tem;
--ch[1][S[first++]];
--cnt;
}
}
return s;
}
};
原文：http://www.cnblogs.com/liyangguang1988/p/3954273.html