java 程序，用于查找作为输入提供的两个整数之间的所有回文完全平方

Question

我需要编写一个程序来查找作为输入提供但不包括提供的输入的两个整数之间的所有回文完全平方。

当您输入大范围的输入（即 10000 和 100000）时，我的程序得到 "killed"。请问我该如何解决这个问题？

import static java.lang.Math.sqrt;
import java.util.Scanner;

public class PalinPerfect {

    public static void main(String[] args){

        Scanner user_input = new Scanner(System.in);
        String start_point;

        System.out.print("Enter start point N:\n");

        start_point = user_input.next();
        int start = Integer.parseInt(start_point);

        String finish_point;
        System.out.print("Enter ending point M:\n");

        finish_point = user_input.next();
        int finish = Integer.parseInt(finish_point);
        System.out.print( "The palindromic perfect squares are as follows:\n");

        for(int i=start; i <=finish; i++){

            int a,n1,n2=0;
            n1=i;

            for(a = 0; a <= i; a++) {

                if (a==Math.sqrt(n1)) {
                    n2=n1;
                }
            }
            int number = n2;
            int reversedNumber = 0;
            int temp=0;
            while(number > 0){
                temp = number % 10;
                number = number / 10;
                reversedNumber = reversedNumber * 10 + temp;
            }
                if(n1 == reversedNumber)
                    System.out.println(n1);

        }
    }
}

Answer 1

sqrt 是一项昂贵的操作，您经常使用它。一种不同的方法是做完全相反的事情：计算数字的平方，如果它在范围内，看看它是否是回文。

当然，这种方法运行得更快：

public static void main(String[] args){

    Scanner user_input = new Scanner(System.in);
    String start_point;

    System.out.print("Enter start point N:\n");

    start_point = user_input.next();
    int start = Integer.parseInt(start_point);

    String finish_point;
    System.out.print("Enter ending point M:\n");

    finish_point = user_input.next();
    int finish = Integer.parseInt(finish_point);
    System.out.print( "The palindromic perfect squares are as follows:\n");

    for(int i = (int)Math.ceil(Math.sqrt(start)); i * i <= finish; i++){
        int number = i * i;
        int reversedNumber = 0;
        int temp = 0;
        while(number > 0){
            temp = number % 10;
            number = number / 10;
            reversedNumber = reversedNumber * 10 + temp;
        }
        if(i * i == reversedNumber)
            System.out.println(reversedNumber);
    }
}

Answer 2

这是另一种简单明了的回文平方数打印方法。

Make sure you check isPalindrome operation first then only you check isPerfectSquare.

System.out.print("The palindromic perfect squares are as follows:\n");

        for (int i = start + 1; i < finish; i++) {
            if (isPalindrome(i) && isPerfectSquare(i) ) {
                System.out.println(i);
            }

        }

定义 isPerfectSquare 和 isPalindrome 方法如下

static boolean isPerfectSquare(int input) {
    int SquareRoot = (int) Math.sqrt(input);
    return ((SquareRoot * SquareRoot) == input);
}

static boolean isPalindrome(int input) {
    String str = Integer.toString(input);
    return new StringBuffer(str).reverse().toString().equals(str);
}

java 程序，用于查找作为输入提供的两个整数之间的所有回文完全平方

java program that finds all palindrome perfect squares between two integers supplied as input

java

palindrome

perfect-square