#include <iostream>
#include <string>
#include <stdio.h>
#include <time.h>

using namespace std;

void PrintCharArray(string a);
bool CharContain(string src, string des);

int main(int argc, char *argv[])
{
  /*
  *  字符串的移位包含问题：给定两个字符串s1和s2，要求判断s2是否能够被s1做循环移位(rotate)得到的字符串包含。
  *  eg: s1 = AABCD, s2 = CDAA, 返回true; s1 = ABCD, s2 = ACBD, 返回false.
  *  注意s1循环移位所有可能包含于s1s1中，因此，可以用string的find方法判断s1s1中是否包含s2.
  */

  //string src = "AABBCD";
  //string des = "CDAA";
  //PrintCharArray(src);
  //PrintCharArray(des);
  //CharContain(src, des);

  clock_t start_time = clock();

  CharContain("AABBCD", "CDAA");
  CharContain("ABCD", "ACBD");

  clock_t end_time = clock();

  cout << "Running time of CharContain1 is " 
    << static_cast<double>(end_time - start_time) / CLOCKS_PER_SEC * 1000
    << "ms." << endl;
  
  return 0;
}

void PrintCharArray(string a)
{
  if (!a.empty())
  {
    int len = a.size();
    for (int i = 0; i < len; i++)
    {
      cout << a[i];
    }
  }
  cout << endl;
}

bool CharContain(string src, string des)
{
  int len = src.size();
  string tmp = src + src;

  int index = tmp.find(des);

  if (index != -1)
  {
    cout << "Case One: " << endl;
    for (int i = index; i < len * 2; i++)
    {
      cout << tmp[i];
    }
    cout << endl;
    cout << "s2 in s1s1." << endl << endl;
    return true;
  }
  else
  {
    cout << "Case Two: " << endl;
    cout << "s2 not in s1s1." << endl;
    return false;
  }
}


字符串循环移位包含问题_面试