Minimum Window Substring -- LeetCode

题目：

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：使用两个指针，一个hash表，两个指针为了记录宽度，hash表为了记录是否存在，目标字符在两个指针之间出现了多少次，直到所有字符都出现在了这段子串内，移动前面的指针，直到某一个字符出现在子串中一次，那么这个空间的长度就是当前最短的，这样遍历直到把所有字符串遍历结束

#include <iostream>
#include <vector>
#include <string>
using namespace std;
/*
一个字符串能包含另一个字符串中所有字母的子串的最小长度 
*/ 

string MinLength(string& src,string& dest)
{
	int i=0,j=0;
	int flag =0;
	int len=src.size();
	int pos=0;
	vector<int> hash(26,-1);
	for(i=0;i<dest.length();i++)
		hash[dest[i]-'A'] =0;
//	for(i=0;i<hash.size();i++)
//		cout<<hash[i]<<endl;
	for(i=0;i<src.size();i++)
	{
		if(hash[src[i]-'A'] >=0)
		{
			hash[src[i]-'A']++;
			if(hash[src[i]-'A'] ==1)
				flag++;
		
		}
		if(flag == dest.length())
		{
			//cout<<"=="<<endl;
	//		cout<<"j is "<<j<<" i is "<<i<<endl;	
			for(;j<i;j++)
			{
				if(hash[src[j]-'A'] == 1)
					break;
				else
					hash[src[j]-'A']--;
					
			}
			if(len >i-j+1)
			{
				len = i-j+1;
				pos =j;						
			}
				
			hash[src[j]-'A'] = 0;
			j++;
			flag--;
		}
	}
	cout<<string(src,pos,pos+len)<<" pos is "<<pos<<" len is "<<len<<endl;;
	return string(src,pos,pos+len);
} 

int main()
{
	string src("ADOBECODEBANCAAABC");
	string dest("ABC");
	cout<<MinLength(src,dest);
	return 0;
}