如何获取整数的位数并将其存储到数组中？ [关闭]

Question

我对编程很陌生。我在 Programming Praxis 中发现了这个有趣的问题，并被困在如何获取整数的数字并将它们存储到数组中。

老实说，我不知道从哪里开始，我想学习如何在 Java 中做到这一点。

如果您有兴趣，这是我正在处理的问题的代码。

public class MyClass {
    public static void main(String args[]) {
    // Determine all three-digit numbers N having the property that N is divisible by 11, and N/11 is equal to the sum of the squares of the digits of N.
    int num = 100;
    int square_sum = 0;
    int digit1 = 0;
    int digit2 = 0;
    int digit3 = 0;
    while (num <= 999){
        
        if (num % 11 == 0) { //if remainder == 0; the number is divisible by 11// 
           
            // We need to get the digits of int "num" and square them // 
            int arrayDigits[] = new int[3];

            } else 
                num++;
        }
    }
}

Answer 1

这是一个示例：（与您的代码不匹配，因此您必须考虑一下）

 int[] arr = new int[] {100, 123, 21};

//get first index of your array
        int a = arr[0];

//turn it into a string
        String b = String.valueOf(a);

//split the string
        String[] c = b.split("");

        System.out.println(c[0]);

通过这种方式，您的String[] c 是一个字符串数组，您可以像访问普通数组一样访问这些值

Answer 2

public static void main(String args[]) {
    // Determine all three-digit numbers N having the property that N is divisible by 11, and N/11 is equal to the sum of the squares of the digits of N.
    int num = 100;
    int square_sum = 0;
    int digit1 = 0;
    int digit2 = 0;
    int digit3 = 0;
    while (num <= 999) {
        if (num % 11 == 0) { //if remainder == 0; the number is divisible by 11//
            // We need to get the digits of int "num" and square them //
            int arrayDigits[] = new int[3];
            digit1 = num % 10;
            digit2 = (num / 10) % 10;
            digit3 = (num / 100) % 10;
            square_sum = digit1 * digit1 + digit2 * digit2 + digit3 * digit3;
            if ((num / 11) == square_sum) {
                System.out.println(num);
            }
        }
        num++;
    }
}

如果你想得到 int "num" 的每个数字

digit1 = num % 10;
digit2 = (num / 10) % 10;
digit3 = (num / 100) % 10;