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【LeetCode】7 - Reverse Integer

时间:2015-07-27 22:37:47      阅读:189      评论:0      收藏:0      [点我收藏+]

Reverse digits of an integer.

Example1: x = 123, return 321
Example2: x = -123, return -321

notice

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?

For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

Update (2014-11-10):
Test cases had been added to test the overflow behavior.

 1  int reverse(int x)   
 2     {  
 3         long long num=x;    //必须先将int提升为long long,否则-INT_MIN会溢出
 4         long long result=0;
 5         int sign=1;
 6         if(num<0){
 7             sign=-1;
 8             num=-num;
 9         }
10         while(num!=0){
11             result=result*10+num%10;
12             num/=10;
13         }
14     
15         if(result>INT_MAX){
16             if(result==INT_MAX+1&&sign==-1)return INT_MIN;
17             else
18                 return 0;
19         }
20         return result*sign;
21     }  

 

【LeetCode】7 - Reverse Integer

原文:http://www.cnblogs.com/irun/p/4681332.html

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