设计模式(15) 迭代器模式（Iterator）C++实现

题目：

Assume you have a method isSubstring which checks if one word is a substring of another. 

Given two strings, s1 and s2, write code to check if s2 is a rotation of s1 using only one call to isSubstring

 (i.e., “waterbottle” is a rotation of “erbottlewat”).

解决思路：

1. Check if length(s1) == length(s2). If not, return false.
2. Else, concatenate s1 with itself and see whether s2 is substring of the result.

• input: s1 = apple, s2 = pleap ==> apple is a substring of pleappleap
• input: s1 = apple, s2 = ppale ==> apple is not a substring of ppaleppale

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

bool isSubstring(string s1, string s2){
    if(s1.find(s2) != string::npos) return true;
    else return false;
}
bool isRotation(string s1, string s2){
    if(s1.length() != s2.length() || s1.length()<=0)
        return false;
    return isSubstring(s1+s1, s2);
}

int main(){
    string s1 = "apple";
    string s2 = "pleap";
    cout<<isRotation(s1, s2)<<endl;
    cout<<string::npos<<endl;
    return 0;
}

设计模式(15) 迭代器模式（Iterator）C++实现,布布扣,bubuko.com

设计模式(15) 迭代器模式（Iterator）C++实现

原文：http://blog.csdn.net/zs634134578/article/details/21232937