求一个字符串中连续出现次数最多的子串

#include "stdafx.h"#include<iostream>#include<vector>#include<string>using namespace std;/*基本算法描述:    给出一个字符串abababa     1.穷举出所有的后缀子串        substrs[0] = abababa;        substrs[1] = bababa;        substrs[2] = ababa;        substrs[3] = baba;        substrs[4] = aba;        substrs[5] = ba;        substrs[6] = a;    2.然后进行比较        substrs[0]比substrs[1]多了一个字母,如果说存在连续匹配的字符,那么        substrs[0]的第1个字母要跟substrs[1]首字母匹配,同理        substrs[0]的前2个字母要跟substrs[2]的前2个字母匹配(否则不能叫连续匹配)        substrs[0]的前n个字母要跟substrs[n]的前n个字母匹配.        如果匹配的并记下匹配次数.如此可以求得最长连续匹配子串.     */pair<int,string> fun(const string& str){vector<string> substrs;int maxcount =1,count =1;string substr;int i,len =str.length();for(i=0;i<len;++i)substrs.push_back(str.substr(i,len-i));for(i=0; i<len; ++i)        cout << substrs[i] << endl;for(i=0;i<len;++i){for(int j=i+1;j<len;++j){count =1;if(substrs[i].substr(0,j-i)==substrs[j].substr(0,j-i)){++count;for(int k=j+(j-i);k<len;k +=j-i)//步长为j-i，子串的长度{if(substrs[i].substr(0,j-1)==substrs[k].substr(0,j-1))++count;elsebreak;}if(count>maxcount){maxcount = count;substr=substrs[i].substr(0,j-1);}}}}return make_pair(maxcount,substr);}int _tmain(int argc, _TCHAR* argv[]){string str="abcbcbcabc";pair<int,string> rs = fun(str);}