題目：

Given a 32-bit signed integer, reverse digits of an integer.

Example 1:

Input: 123
Output:  321

Example 2:

Input: -123
Output: -321

Example 3:

Input: 120
Output: 21

Note:
Assume we are dealing with an environment which could only hold integers within the 32-bit signed integer range. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

這個題目是一個easy的題目，但是細節要註意，第一點要註意記得處理負數的情況，

第二點就是記得看Note,Assume we are dealing with an environment which could only hold integers within the 32-bit signed integer range. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

就是超過int的範圍了就需要返回0。這一點非常重要，因為你要判斷的是可能反轉後會溢出，那麽需要註意的就是

需要用一個long long來存儲答案ans。判斷使用的語句就是：

if(ans>INT_MAX||ans<INT_MIN)
return 0;

啦啦啦！

代碼如下：

 1 class Solution {
 2 public:
 3     int reverse(int x) {
 4         long long ans=0;
 5         bool isLittle=false;
 6         if(x<0)
 7         {
 8             x=-x;
 9             isLittle=true
 
;
10         }
11         while(x>0)
12         {
13             int num=x%10;
14             ans=ans*10+num;
15             x=x/10;
16         }
17         if(ans>INT_MAX||ans<INT_MIN)
18             return 0;
19         if(isLittle)
20             return -ans;
21         else return ans;
22     }
23 };

leetcode題解 7.Reverse Integer