leetcode007. Reverse Integer

 /* a good way to predict overflow
  * each time *10 must predict int overflow
  * not only the last time to predict which will be wrong.
  */
 class Solution {
 public:
     int reverse(int x) {
         if(x==)return x;
         int tag=;
         long int re=;
         if(x<)
         {
             tag=;
             x=-x;
         }
         while(x)
         {
             int tmp=re*+x%;
             if((tmp-x%)/!=re)return ;
             re=tmp;
             x=x/;
         }
         if(tag)re=-re;
         return re;
     }
 };