题解 | #回文数判断#

使用StringBuilder拼接字符串，可拓展性比较强，不必拘泥于数字位数

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner console = new Scanner(System.in);
        Main main = new Main();
        int number = console.nextInt();
        System.out.println(main.palindromeNumber(number));
    }

    public Boolean palindromeNumber(int number) {
        StringBuilder str=new StringBuilder();
        str.append(number);
        int left=0;
        int right=str.length()-1;
        while(left<right){
            if(str.charAt(left)!=str.charAt(right)){
                return false;
            }else{
                left++;
                right--;
            }
        }
        return true;
    }
}