【LeetCode】Reverse digits of an integer

Reverse digits of an integer.

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

Have you thought about this?

Here are some good questions to ask before coding. Bonus points for you if you have already thought through this!

If the integer's last digit is 0, what should the output be? ie, cases such as 10, 100.

Did you notice that the reversed integer might overflow? Assume the input is a 32-bit integer, then the reverse of 1000000003 overflows. How should you handle such cases?

For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

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

 

java中取绝对值的函数：Math.abs(变量名)

对于溢出的判断。可以直接与Integer.MIN_VALUE和Integer.MAX_VALUE比较。也可以自己初始化min_value = 0x80000000,max_value= 0x7fffffff。

if语句尽量用条件运算语句代替，可以简化代码，使代码逻辑更加清晰。

package com.jie.easy;

import java.util.Scanner;

public class ReverseInteger {
    public static void main(String []args){
        Scanner sc = new Scanner(System.in);
        System.out.println("input:");
        int a = sc.nextInt();
        int result = reverse(a);
        System.out.println("output:\n"+result);
        
    }
    public static int reverse(int x){
        if(x<=Integer.MIN_VALUE || x>=Integer.MAX_VALUE)
            return 0;
        int res = 0;

//        int flag = 1 ;
//        if(x<0){
//            flag = -1;
//            x = -x;
//        }        
        int flag = x < 0 ? -1 : 1;
        x = Math.abs(x);
        while(x>0){
            res = res * 10 + x % 10;
            x/=10;
        }
        
        return flag*res;
    }
}