字符串的最小窗口问题

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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：

0 0