https://leetcode.com/problems/reverse-integer/description/
輸入: 32位有符號 int
輸出: 逆序
思路:
循環(huán)取余就可以實現(xiàn)了,需要注意溢出的情況

class Solution {
    public int reverse(int x) {
        int t = x;
        int num = 0;
        while(t!=0){
            int temp = num*10 + t%10;
            if((temp-t%10)/10 != num){
                return 0;
            }
            num = temp;
            t=t/10;
        }
        return num;
    }
}