7. Reverse Integer - LeetCode

LeetCode Problems Solutions

question description: 问题描述

Reverse digits of an integer.
The input is assumed to be a 32-bit signed integer. Your function should return 0 when the reversed integer overflows.
将一个整数反向。
输入假定为一个32位的带符号整数。当反向的整数溢出时，函数应该返回0

Example:例如

Example1: x = 123, return 321
Example2: x = -123, return -321

Thinking

看到题目后最直接的思路可能是利用数组进行反向，然后将字符串转为int，最后判断溢出的问题。
还有另一种方法：

solution with java - Java解决方案

public static int reverse(int x) {

        long tmp = x;
        long ret = 0;
        while (tmp != 0)
        {
            ret = ret * 10 + tmp % 10;
            tmp /= 10;
        }

        if (ret > Integer.MAX_VALUE || ret < Integer.MIN_VALUE) {
            return 0;
        }


        return (int)ret;
    }

solution with swift - swift解决方案

func reverse(_ x: Int) -> Int {
        
    var tmp = x
    var ret = 0
    while (tmp != 0)
    {
        ret = ret * 10 + tmp % 10;
        tmp /= 10;
    }
        
        
    if ret > Int(Int32.max) || ret < Int(Int32.min) {
        return 0;
    }    
        
    return ret;
    }