[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.

原文链接：http://blog.csdn.net/jellyyin/article/details/10313827

思路：双指针，动态维护一个区间。尾指针不断往后扫，当扫到有一个窗口包含了所有T的字符后，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

代码：

class Solution {
public:
string minWindow(string S, string T) {
int slen=S.size();
int tlen=T.size();
if(tlen>slen || tlen==0 || slen==0) return "";
int needFind[256]={0};
int hasFind[256]={0};
for(int i=0;i<tlen;i++){
needFind[T[i]]++;
}
int minLength=INT_MAX;
int minbegin=0;
int minend=slen-1;
int begin=0;
int end=0;
for(int count=0;end<slen;end++){
if(needFind[S[end]]==0) continue;
hasFind[S[end]]++;
if(hasFind[S[end]]<=needFind[S[end]]) count++;
if(count==tlen){
while(begin<end){
if(needFind[S[begin]]==0){
begin++;
continue;
}
if(hasFind[S[begin]]>needFind[S[begin]]){
hasFind[S[begin]]--;
begin++;
continue;
}
else
break;
}
int tempLength=end-begin+1;
if(tempLength<minLength){
minbegin=begin;
minend=end;
minLength=tempLength;
}
}
}
if(minLength==INT_MAX) return "";
return S.substr(minbegin,minLength);
}
};