子字符串的最小窗口问题

给定两个字符串T和S，找出在S中包含字符串T的最小子字符串，要求时间复杂度为O（n）。

如果不存在最小窗口，则返回”“；

例如,

S = "ADOBECODEBANC"

T = "ABC"

返回的最小窗口是"BANC".

实验的代码如下所示：

#include<iostream>
#include<string>
#include<vector>
using namespace std;
string minWindow(string &S, string &T)
{
	int Slen = S.length();
	int Tlen = T.length();
	if (Slen < Tlen || T.empty())
	{
		return "";
	}
	vector<int> ToBeFind(128, 0);
	vector<int> Found(128, 0);
	for (int i = 0; i != Tlen; ++i)
	{
		ToBeFind[T[i]]++;
	}
	int Count = Tlen;
	int first = 0;
	int last = 0;
	int minfirst = 0;
	int minlast = 0;
	int minlength = INT_MAX;
	Found[S[0]]++;
	if (Found[S[0]] <= ToBeFind[S[0]])//先比较第一个字符是为了处理T为单个字符的特例
	{
		Count--;
	}
	while (true)
	{
		if (Count == 0)
		{
			while (Found[S[first]] > ToBeFind[S[first]])
			{
				Found[S[first]] --;
				first++;
			}
			int length = last - first + 1;
			if (length < minlength)
			{
				minfirst = first;
				minlast = last;
				minlength = length;
			}
		}
		if (last < Slen)
		{
			last++;
			Found[S[last]]++;
			if (Found[S[last]] <= ToBeFind[S[last]])
			{
				Count--;
			}
		}
		else
		{
			break;
		}
	}
	if (minlength == INT_MAX)
	{
		return "";
	}
	return S.substr(minfirst, minlength);
}
int main()
{
	string S1 = "ADOBECODEBANC";
	string S2 = "ABC";
	string S3 = "A";
	cout << minWindow(S1, S2) << endl;
	cout << minWindow(S3, S2) << endl;
	system("pause");
	return 0;
}

实验结果如下所示：