7. Reverse Integer

 1 class Solution {
 2 public:
 3     int reverse(int x) {
 4         int max =  2147483647;  //int最大值
 5         int min = -2147483648; //int最小值
 6         
 7         long long ret = 0;
 8         while(x!=0){
 9             int temp = x % 10;
10             ret = ret*10 + temp;
11             
12             //溢出处理
13             if(ret > max || ret < min){
14                 return 0;
15             }
16             
17             x = x /10;
18             
19         }
20         return (int)ret;
21     }
22 };