LeeCode整数 反转

用c++写的时候注意处理溢出。

class Solution {
public:
    int reverse(int x) {
        long long result=0;
        int divided, reminder;
        divided = x;
        reminder = 0;
        while(divided) {
            reminder = divided%10;
            divided = divided/10;
            result = result*10 + reminder;
        }
        //处理溢出情况 
        if(result >= -pow(2,31) && result <= pow(2, 31)-1) { 
            return int(result);
        }
        return 0;
    }
};