编程之美：3.1 字符串移位包含的问题

#include<iostream>
#include<string>
using namespace std;

/*
该函数主要是遍历的方法，将所有情况都测试一遍，效率较低
时间复杂度为O(n2)
*/
void function1(string src, string des)
{
int len = src.length();
//src += "123";
//cout << src << endl;
for (int i = 0; i < len; i++)
{
char temp = src[0];
for (int j = 0; j < len - 1; j++){
src[j] = src[j + 1];
}
src[len -1] = temp;

//判断是否包含该字符串
if (src.find(des) != string::npos){
cout << "true" << endl;
return;
}
}
}
/*
This function is based on the theory that if s2 can roll from
s1, then s2 must be a substr of s1s1.
*/
void function2(string src, string des)
{
src += src;
cout << src << endl;

if (src.find(des) != string::npos){

cout << "true" << endl;
return;
}

return;
}
int main()
{
string src;
string des;

while (cin >> src){
cin >> des;

//在function1中对src 和 des的改变将不会影响主程序的src和des,这和java有一定的区别
//function1(src, des);
//cout << src << endl;
function2(src, des);
}

}