Question

我需要编写一个程序，找到作为输入提供的两个整数之间的所有回文完美正方形，但不包括提供的输入。

当您输入大量输入时，我的程序会被“杀死”，即10000和100000.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，而且您经常使用它。另一种方法是完全相反：计算数字的平方，如果它在范围内，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

这是打印回文平方数的另一种简单而简洁的方法。

确保首先检查isPalindrome操作，然后才检查 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＆amp; 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);
}

找到作为输入提供的两个整数之间的所有回文完美正方形的java程序

2 个答案: