LeetCode: Minimum Window Substring

[b]LeetCode: Minimum Window Substring[/b]

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.

地址：https://oj.leetcode.com/problems/minimum-window-substring/

算法：由于申请两个大小为256的数组，用于统计每一个字符串中各字母的个数，然后用isWindow函数来判断是否由hashS表示的字符串包含由hashT表示的字符串。首先，从S开始找到第一在T中出现的字母位置，记为i，初始化start为i，其中count表示在字符串S[start~i]之间出现在字符串T的字符个数。在每次进入循环时都判断S[start~i]是否可以包含T，如果包含，则根据情况更新最小窗口，并且更新start为下一个出现在字符串T中位置（因为最小的窗口的其实位置一定会出现在字符串T中），注意还要将改变hashS让其表示当前的S[start~i]（注意i还是不变）；如果不包含，则改变i为下一个出现在字符串T中的位置，并且同时更新对应的hashS。代码：

class Solution {
public:
string minWindow(string S, string T) {
int slen = S.size();
int tlen = T.size();
if(slen <= 0 || tlen <= 0)   return string();
vector<int> hashT(256,0);
vector<int> hashS(256,0);
for(int i = 0; i < tlen; ++i)
++hashT[T[i]];
int minNum = slen + 1;
int minStart = 0, minEnd = 0;
int i = 0;
while(i < slen && !hashT[S[i]]){
++i;
}
if(i >= slen)   return string();
int start = i;
int count = 1;
++hashS[S[i]];
while(i < slen){
if(isWindow(hashT,hashS,count,tlen)){
if(minNum > i-start+1){
minNum = i-start+1;
minStart = start;
minEnd = i;
}
--count;
--hashS[S[start]];
++start;
while(start<slen && !hashT[S[start]])   ++start;

}else{
++i;
while(i<slen && !hashT[S[i]])   ++i;
if(i < slen){
++count;
++hashS[S[i]];
}
}
}
if(minNum == slen + 1){
return string();
}
return S.substr(minStart,minNum);
}
bool isWindow(vector<int> &hashT, vector<int> &hashS, int count ,int tlen){
if(count < tlen)
return false;
for(int i = 0; i < 256; ++i){
if(hashS[i] < hashT[i]){
return false;
}
}
return true;
}
};