LeetCode:Reverse Integer

本文介绍了一种用于反转整数的算法实现,通过示例说明了如何处理正负数及潜在的溢出问题,并提供了一个C++代码示例。

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)


leetcode的测试样例中似乎没有翻转后溢出的情况。                      本文地址

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class  Solution {
public :
     int  reverse( int  x) {
         bool  isPositive = true ;
         if (x < 0){isPositive = false ; x *= -1;}
         long  long  res = 0; //为了防止溢出,用long long
         while (x)
         {
             res = res*10 + x%10;
             x /= 10;
         }
         if (res > INT_MAX) return  isPositive ? INT_MAX : INT_MIN;
         if (!isPositive) return  res*-1;
         else  return  res;
     }
};





本文转自tenos博客园博客,原文链接:http://www.cnblogs.com/TenosDoIt/p/3735866.html,如需转载请自行联系原作者

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