leetcode：Reverse digits of an integer

//Reverse digits of an integer.//Example1: x = 123, return 321//Example2: x = -123, return -321//Have you thought about this?//Here are some good questions to ask before coding. Bonus points for you if you have already thought through this!//If the integer's last digit is 0, what should the output be? ie, cases such as 10, 100.//Did you notice that the reversed integer might overflow? Assume the input is a 32-bit integer, then the reverse of 1000000003 overflows. How should you handle such cases?//Throw an exception? Good, but what if throwing an exception is not an option? You would then have to re-design the function (ie, add an extra parameter).using namespace std;#include "Solution12.h"//#include <stack>Solution12::Solution12(){}Solution12::~Solution12(){}//方法1：//int Solution12::reverse(int x)//{//if (x<0)//{//x=abs(x);//}//int max_int=0x7fffffff;//int min_int=0x80000000;//int sum=0;//while (x!=0)//{//int temp=x%10;//sum=sum*10+temp;//if (sum > max_int || sum< min_int)//{//sum=sum>max_int?max_int:max_int;//return sum;//}//x=x/10;//}//return sum;//}//方法二：class Solution {  public:  int reverse(int x) { int sign = x < 0? -1: 1;  int ret = 0;  x = abs(x);  while(x > 0)  {  ret = ret*10 + x%10;  x /= 10;  }  return sign*ret;  }  };  

两种方法都AC，道理差不多。