LeetCode[7]Reverse Integer

Reverse digits of an integer.

Example1: x = 123, return 321

Example2: x = -123, return -321

Note:
The input is assumed to be a 32-bit signed integer. Your function should return 0 when the reversed integer overflows.

本地可执行程序：

#include<iostream>using namespace std;/*解题思路：把数字对10取余，在乘10逐渐累加，最后判断溢出*/int reverse(int x);int main(){int result;int a = -12345006;result = reverse(a);cout << result << endl;system("pause");return 0;}int reverse(int x){long long sum = 0;//反转while (x != 0){int s = x % 10;sum = sum * 10 + s;x = x / 10;}//判断溢出if (sum > INT_MAX || sum < INT_MIN){return 0;}else{return sum;}}

提交代码：

class Solution {public:    int reverse(int x)     {        long long sum=0;        while(x!=0)        {            int s = x%10;            sum = sum*10 + s;            x = x/10;        }        if(sum > INT_MAX || sum < INT_MIN)        {            return 0;        }        else        {            return sum;        }            }};