LeetCode 7. Reverse Integer(数学)

• By Scarb
• Scarb’s Blog

Tags:
- Math

问题描述

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.

解题思路

题意是将一个整数转置输出。提示：转置输出的数溢出时返回0.
所以要处理一下转置后的数溢出的特殊情况，这里包括了正溢出和负溢出。

参考代码

#include <iostream>using namespace std;class Solution {public:    int reverse(int x) {        int num = 0;        while (x)        {            if (abs(num) > INT_MAX / 10 ||              // 处理溢出                (abs(num) == INT_MAX / 10 && abs(x) % 10 > INT_MAX % 10))                return 0;            num = num * 10 + x % 10;            x /= 10;        }        return num;    }};int main(){    int i = 123;    Solution sl = Solution();    cout << sl.reverse(123) << endl;    cout << sl.reverse(-123) << endl;    cout << sl.reverse(103) << endl;    cout << sl.reverse(120) << endl;    cout << sl.reverse(1000000003) << endl;    cout << sl.reverse(-1000000003) << endl;    system("pause");    return 0;}