Java 数字回文

Question

我想写一个回文程序，打印所有由两位数 (10-99) 相乘得到的回文数？

到目前为止，这是我的代码：

public class PrintPalindrom {
    public int printPalindrom (int a, int b) {
        int result = a*b;
        int reverse = 0;
        if (a >= 10 && a <= 99 && b >= 10 && b <= 99) {
            while (result != 0) {
                reverse = reverse * 10;
                reverse = reverse + result % 10;
                result = result/10;
                System.out.println("palindrom is " + result);
            }
        } else {
            System.out.println("Wrong numbers");
        }
        return result;
    }
}

如何打印两位数(10-99)相乘产生的回文数？

我的方法暂时有效...

Answer 1

创建一个方法 (isPalindrome()) returns 如果数字是回文数则为 true，否则为 false。您可以从 printPalindrom() 方法开始轻松创建它。然后，在你的 main():

int from = 10;
int to = 99;
for (int i = from; i <= to; i++) {
    for (int j = from ; j <= to; j++) {
        int mult = i*j;
        if (isPalindrome(mult))
            System.out.println(i + "x" + j + "=" + mult + " is plaindrome!");
    }
}

Answer 2

请检查此代码：

 public class PrintPalindrom {
    public int printPalindrom (int a, int b) {
        int result = a*b;
        int temp = result; 
        int reverse = 0;
        if (a >= 10 &&  a <= 99 && b >= 10 && b <= 99) {
            while (temp != 0) {
                reverse = reverse * 10;
                reverse = reverse + result % 10;
                temp = temp/10;                
            }
           if(result == reverse) {            
            return result;   
           } 
       }
      return 0;     

    }   

     int from = 10;
     int to = 99;
     for (int i = from; i < to; i++) {
         for (int j = i + 1 ; j <= to; j++) {
              int result = printPalindrom(i, j);
              if(result > 0) {
                 System.out.println(result + ",");
             }              
        }
    }
}

Answer 3

你可以试试这个，如果允许使用堆栈来检查两个数字的乘积是否是回文。

import java.util.Stack;

public class PrintPalindrom {
    public static void main(final String[] args) {
        final int MIN = 10;
        final int MAX = 99;
        for (int i = MIN; i <= MAX; i++) {
            for (int j = MIN ; j <= MAX; j++) {
                new PrintPalindrom(i, j);
            }
        }
    }

    public PrintPalindrom(final int a, final int b) {
        if (a >= 10 &&  a <= 99 && b >= 10 && b <= 99) {
            final String result = String.valueOf(a * b);    //Convert the product of the int to a String
            final Stack<String> stack = new Stack<String>();

            System.out.print(a + " * " + b + " = " + result + "\t-\t");

            //Push each character from the in to a stack
            for(int n = 0; n < result.length(); n++) {
                stack.push(result.substring(n, n + 1));
            }

            //Check each character against the characters in the stack
            for(int n = 0; n < result.length(); n++) {
                if(!result.substring(n, n + 1).equals(stack.pop())) {
                    System.out.println("Not a palindrom");
                    return;
                }
            }
        }

        System.out.println("A palindrom");
    }
}

Answer 4

public class Example {

    public static void main (String[]args){
        int start = 10;
        int end = 99;            
        for (int i = start; i < end; i++) {
            for (int j = i ; j <= end; j++) {
                 printPali(i,j);              
           }
        }
    }

    public static void printPali(int i, int j) {            
        if(String.valueOf(i*j).equals(new StringBuilder(String.valueOf(i*j)).reverse().toString()))
            System.out.println(i + " * " + j + " = " + i*j);
    }
}

Answer 5

你可以通过这种方式实现创建 isPalindrome 函数，其中 return true 如果数字是回文，则检查所需范围内的每个乘法结果的回文，如果这是 true.

 public static boolean isPalindrome(int num){
              int rnum = 0,digit;
              int n = num;
              while(n > 0){
                    digit = n % 10;
                    n = n / 10;
                    rnum = rnum * 10  + digit;
              }
              if(rnum == num){
                  return true;
               }
               else{
                  return false;
               }
 }        
 public static void main(String[] args) {          
            int a = 10 , b = 99;
        for(int i = a; i <= b; i++){
            for(int j = i; j <= b; j++){
                 if(isPalindrome(i * j)){
                     System.out.println("Mult Palindrome: " + i * j);
                 }
            }
        }   
 }

请注意，您可以将范围放在内部 for 循环 for(int j = a; j <= b; j++) 中，但它在乘法结果中是多余的，没有任何好处，因此将范围从 j=i 到b.

希望对您有所帮助。

Answer 6

import java.util.Scanner;

public class Javatips {

    public static void main(String[] args){

        Scanner in = new Scanner(System.in);
        int number = in.nextInt();
        int reverseNumber = 0;
        int copyNumber = number;
        boolean isPalindrom = true;
        while (copyNumber>0){
            reverseNumber = 10*reverseNumber + copyNumber%10;
            copyNumber=copyNumber/10;
            if(reverseNumber==number){
                isPalindrom = true;
            } else {
                isPalindrom = false;
            }
        }
        if(isPalindrom){
            System.out.println(number + " is palindrom");
        } else {
            System.out.println(number + " is not palindrom");
        }
    }
}

Java 数字回文

Java Number Palindrom

java

numbers

palindrome