找到由两个3位数字的乘积制成的最大回文。 （JAVA） [英] Find the largest palindrome made from the product of two 3-digit numbers. (Java)

问题描述

我在解决Project Euler问题4时遇到了一些麻烦。我对编程缺乏经验，并没有真正理解其他答案。我设法写了一个代码，打印所有六位数的回文。如何找到由两个3位数字相乘得出的最大回文？

I am having a bit of trouble solving Project Euler question 4. I am inexperienced in programming and didn't really understand other answers. I managed to write a code that prints all six-digits palindromes. How do I find the largest palindrome made from the multiplying of two 3-digit numbers?

public class Main {

    public static void main(String[] args) {
        int num = 998001;
        int count=0;
        int temp1, temp2, temp3, temp4, temp5, temp6;
        int temp;
        int num1=999;
        int num2=100;
        for (int i = 100000; i <= 998001; i++) {
            num=998001-count;
            temp=num;
            temp1=temp%10;
            temp=temp/10;
            temp2=temp%10;
            temp=temp/10;
            temp3=temp%10;
            temp=temp/10;
            temp4=temp%10;
            temp=temp/10;
            temp5=temp%10;
            temp=temp/10;
            temp6=temp%10;
            temp=temp/10;

            if (temp1==temp6 && temp5==temp2 && temp3==temp4) {
                System.out.println(num);
            }
            count=count+1;
        }
    }
}

推荐答案

这是一种循环遍历1-999的所有数字并将它们与1-999中所有可能数字相乘的方法。您需要定义一个方法 isPalindrome（int），它将检查给定的int是否是回文。

Here is a method which loops over all numbers from 1-999 and multiplies them with all possible numbers from 1-999. You will need to define a method isPalindrome(int) that will check if a given int is a palindrome.

//save the largest number
int largest = 0;

//loop over every possible product of numbers from 100-999
for (int i = 999; i >= 100; i--) {
 for (int j = i; j >= 100; j--) {
  int curr = i * j;

  //if the current number is a palindrome and is greater than the last found largest
  if (isPalindrome(curr) && curr > largest) {

   //save it as the new largest found number
   largest = curr;
  }
 }
}

isPalindrome 方法可能如下所示：

private static boolean isPalindrome(Integer possible) {
    String toStr = possible.toString();
    int len = toStr.length();
    if (len % 2 == 1) {
        len = len - 1;
    }

    for (int i = 0; i < len / 2; i++) {
        if (toStr.charAt(i) != toStr.charAt(toStr.length() - (1 + i))) {
            return false;
        }
    }

    return true;
}

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