【Leetcode】Reverse Integer_整數反轉_Easy2

本文思路參考：

做者：青貓123 
來源：CSDN 
原文：https://blog.csdn.net/superstar987/article/details/80426102 

整數反轉題目：主要是要反轉整數+判斷溢出

Given a 32-bit signed integer, reverse digits of an integer.

Example 1:

Input: 123
Output: 321

Example 2:

Input: -123
Output: -321

Example 3:

Input: 120
Output: 21

Note:
Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−2³¹,  2³¹ − 1]. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

 

這道題的要求是要反轉整數，可是若是反轉事後溢出的話，就要返回0；重點在於如何判斷是否溢出。

 Answer：

class Solution
{
    public static int reverse(int num){
        {
            int length = 0;    //給出數的長度
            int rev = 0;
            if(num >= 0) length = (num+"").length();
            else length = (num+"").length();   //負數會佔用一位符號位
            for(int i = 0; i < length; i++)      
            {
                int temp = num % 10;      //取最後一位數
                num = (num - temp) / 10;   //把最後一位拋掉之後獲得的新的數
                rev += temp*Math.pow(10, length - i -1);    //反轉數
            }
            
            if(rev > Math.pow(2, 32) - 1 || rev < (-1) * Math.pow(2, 31))  //判斷溢出，簡單暴力
                return 0;
            return rev;
        }
    }
}

 

 

public int reverse(int x) {
    long result = 0;
    while (x != 0) {
        result = result * 10 + x % 10;
        x /= 10;   //算法比上一種更加簡單
    }
    if (result > Integer.MAX_VALUE || result < Integer.MIN_VALUE)   //Integer.MAX_VALUE就是232
        result = 0;
    return (int)result;
}