Question

如上所述，我有两种方法，

public static int reverse(int number)

和

public static boolean isPalindrome(int number)

我应该安排他们检查一个号码是否是回文。我试图应用提示（一个数字是回文，如果（它的逆转==本身）），但它没有帮助。

import java.util.Scanner;

public class Test {
    public static void main(String[] args) {

        System.out.println("This program checks whether a number is palindrome or not. Enter a number: ");
        Scanner input = new Scanner(System.in);
        int inputNumber = input.nextInt();

        // if inputNumber is palindrome 

        System.out.println("The number is palindrome.");

        // else 

        System.out.println("The number is not a palindrome.");
    }


        public static int reverse(int number)

        // reverse number


        public static boolean isPalindrome(int number)

        // check if the number is palindrome     
    }
}

Answer 1

为了反转一个数字，你应该将它除以10并将余数加到一个新的反转数字乘以10。

例如，这将是恢复123的步骤：

反转= 0，目标= 123
反转= 0 * 10 + 123％10 = 0 + 3 = 3，目标= 123/10 = 12
反转= 3 * 10 + 12％10 = 30 + 2 = 32，目标= 12/10 = 1
反转= 32 * 10 + 1％10 = 320 + 1 = 321，目标= 1/10 = 0

以下是Java中该算法的实现：

public static int reverse(int number) {
    int reversed = 0;
    while (number > 0) {
        reversed *= 10;
        reversed += number % 10;
        number /= 10;
    }
    return reversed;
}

public static boolean isPalindrome(int number) {
    return number == reverse(number);
}

Answer 2

尝试使用此方法进行反向。

Step0  : temp = 0 number  = 4554
Step 1 : temp = 4 number = 455
Step 2 : temp = 4*10 +  5 = 45 and number = 45
Step 3 : temp = 45*10 + 5 = 455 and number  = 4
Step 4 : temp = 455*10 + 4 = 4554 and number  = 0


    if (inputNumber == reverse(inputNumber)) {
        System.out.println("The number is palindrome.");
    }else{
        System.out.println("The number is not a palindrome.");
    }

public static int reverse(int number) {
    int temp = 0;
    while (number > 0) {
        temp = temp * 10 + number % 10;
        number = number / 10;
    }
    return temp;
}

Answer 3

"Hello - Adele (Cover)" -> "Hello - Adele "
"Flo Rida - Wild Ones ft. Sia [Official Video]" -> "Flo Rida - Wild Ones ft. Sia"
"Hello - Adele (Cover) [Video]"

这是解决问题的方法。但这并不包括我可以进入＆＃34; 011＆＃34;该计划将说＆＃34; 11＆＃34;是一个回文。我怎么能解决这个问题？

Answer 4

以下算法将起作用

1. Loop until the input number become 0.

    1.1 remainder = number % 10;
    1.2 reverse = reverse * 10 + remainder;
    1.3 number = number / 10;

2. Return reverse when the loop breaks. That means number became 0.

Check full implementation here

我想实现两种方法来检查数字是否是回文

4 个答案: