Simple Number to Array with Individual Digits

问题

I am exceptionally new to programming, but I am working on improving my skills as a programmer. Currently, I am working on a problem I gave myself where I am trying to take a variable number and make each of its digits a separate number in an array. I don't care about the order of the digits, so if they are reversed, then it doesn't matter to me. I know people have asked this question numerous times, but they always seem to use a lot of things that I don't understand. Since my school doesn't offer any Java courses, I only know what I have learned on my own, so if you could explain any terms you use in the code that aren't extremely trivial, that would be wonderful. Right now, I have written:

int number = 1234567890;
    while (number > 0) {
        System.out.println(number%10);
        number = number/10;

This works fine for printing the digits individually, but I can't figure out how to add them to the array. I greatly appreciate any help you can give, and please keep in mind that I much prefer simplicity over small size. Thank you in advance!

P.S. Some responses I've seen for similar questions include what I think are arrays of strings. In order for the part of the program that I have working to still work, I think that I need to use an array of integers. If you're curious, the rest of the code is used to determine if the numbers in the array are all different, in order to achieve the final result of determining if a number's digits are all distinct. It looks like this:

int repeats=0;
int[] digitArray;
digitArray = new int[10];
for (int i = 0; i < digitArray.length; i++) 
    for (int j = 0; j < digitArray.length; j++)
        if ((i != j) && (digitArray[i]==digitArray[j])) unique = unique+1;
System.out.println(unique==0);

回答1:

Method number.toString().length() will return the number of digits. That is the same as the length of your needed array. Then you use your code as before, yet instead of printing you add the digit to the array.

int number = 1234567890;
int len = Integer.toString(number).length();
int[] iarray = new int[len];
for (int index = 0; index < len; index++) {
    iarray[index] = number % 10;
    number /= 10;
}

回答2:

I would rather suggest you to use an ArrayList, since to use an array, you would have to allocate the size in advance, for which you need to know the number of digits in your number, which you don't know.

So, either work with an array, and do the iteration over the number twice - once for finding size, and next for doing actual work. Else, move ahead with an ArrayList.

Adding an element to an ArrayList is quite simple. You just need to call - List#add(E) method with appropriate parameter.

So, here's an extension of your solution: -

// Declare a List<Integer>, since it will store integers only.
List<Integer> digits = new ArrayList<Integer>():

int number = 1234567890;
while (number > 0) {
    int digit = number % 10;  // Store digit in a variable
    number = number/10;

    // Add digit to the list
    digits.add(digit);
}

Alternatively, if you want to have only unique digits in your List, then you should use a HashSet, which automatically removes the duplicates.

回答3:

With Java 8:

Integer.toString(n).chars().map(a->a-'0').toArray()

回答4:

char [] arr = scan.next().toCharArray();

This code will read a number from scan.next() and then it is going to give it as an input to char array which will have the number at its indices as single digit by digit.

Hope this will help.

来源：https://stackoverflow.com/questions/14743165/simple-number-to-array-with-individual-digits

标签

java

arrays

digits