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Storing digits of numbers in array

Hello fellow programmers !
I am a beginner with Java and i am looking for a method or a way maybe to store the digits of a 6 digit number entered by the user , in an int array.
For example :-
if the number is 675421.
then i want to store the digits in an array like :-
int[] array = new int[6];
int number = 675421
array[0] = 6;
array[1] = 7;
array[2] = 5;
array[3] = 4;
array[4] = 2;
array[5] = 1;
I want to do so so that i can work with the array to maybe sort or change the order or numbers in array. Thanks!

Here you go,
String temp = Integer.toString(number);
int[] num = new int[temp.length()];
for (int i = 0; i < temp.length(); i++){
num[i] = temp.charAt(i) - '0';
}
for (int i = 0; i < temp.length(); i++) {
System.out.println(num[i]);
}
Edit, after comment
Here, First, you are converting to your number to a string.
Then, take each char out of it(in the loop), subtract the ASCII value of 0 from each char to get the digit [ie, ASCII of 0 is 48, 1 is 49, ... ] (see ASCII table)

Do something like this:
String number = "123123";
int[] intArray = new int[number.length()];
for (int i = 0; i < number.length(); i++)
{
intArray[i] = Integer.parseInt(Character.toString(number.charAt(i)));
}
Hope this helps,
Jason.

Below is the recursive solution
public static void main(String[] args) {
int testNum = 675421;
List<Integer> digitList = new ArrayList<Integer>();
collectDigits(testNum, digitList);
Object[] resultArr = digitList.toArray();
int listSize = resultArr.length;
for (int listCount = 0; listCount < listSize; listCount++) {
System.out.println("result["+listCount+"] = "+resultArr[listCount]);
}
}
private static void collectDigits(int num, List<Integer> digits) {
if (num / 10 > 0) {
collectDigits(num / 10, digits);
}
digits.add(num % 10);
}

One way to do this would be to turn the original integer into a string.
Loop over the string, parsing each character back to an int, and place into the array. Here is an example:
int number = 123456;
String strNumber = number+"";
int[] array = new int[strNumber.length()];
int index = 0;
for(char c : strNumber.toCharArray()){
array[index++] = Integer.parseInt(c+"");
}
System.out.println(Arrays.toString(array));

Math solution, you can split the int number using this:
int[] array = new int[6];
int number = 675421;
array[0] = ((number/100000)%10);
array[1] = ((number/10000)%10);
array[2] = ((number/1000)%10);
array[3] = ((number/100)%10);
array[4] = ((number/10)%10);
array[5] = ((number/1)%10);
If the "number" has a variable length you can automate this, write a coment if you need help

How do I fill an array with consecutive numbers

I would like to fill an array using consecutive integers. I have created an array that contains as much indexes as the user enters:
Scanner in = new Scanner(System.in);
int numOfValues = in.nextInt();
int [] array = new int[numOfValues];
How do i fill this array with consecutive numbers starting from 1?
All help is appreciated!!!

Since Java 8
// v end, exclusive
int[] array = IntStream.range(1, numOfValues + 1).toArray();
// ^ start, inclusive
The range is in increments of 1. The javadoc is here.
Or use rangeClosed
// v end, inclusive
int[] array = IntStream.rangeClosed(1, numOfValues).toArray();
// ^ start, inclusive

The simple way is:
int[] array = new int[NumOfValues];
for(int k = 0; k < array.length; k++)
array[k] = k + 1;

for(int i=0; i<array.length; i++)
{
array[i] = i+1;
}

You now have an empty array
So you need to iterate over each position (0 to size-1) placing the next number into the array.
for(int x=0; x<NumOfValues; x++){ // this will iterate over each position
array[x] = x+1; // this will put each integer value into the array starting with 1
}

One more thing. If I want to do the same with reverse:
int[] array = new int[5];
for(int i = 5; i>0;i--) {
array[i-1]= i;
}
System.out.println(Arrays.toString(array));
}
I got the normal order again..

Scanner in = new Scanner(System.in);
int numOfValues = in.nextInt();
int[] array = new int[numOfValues];
int add = 0;
for (int i = 0; i < array.length; i++) {
array[i] = 1 + add;
add++;
System.out.println(array[i]);
}

Extracting indivdual digits from a number.

I am trying to extracting a individual number from a given integer. example from 1234, I want to store 1 , 2 ,3 ,4 in an array.The number of digits might not be same every time. I don't know how to initialize the array for the same.
int number = 1234;
int [] a = new int[];
for(int i =0;i<lengthOfNum;i++){
a[i] = digitReturn();
}

You can try the following:
int number = 1234;
int length = Integer.toString(number).length();
int[] a = new int[length];
int index = length - 1;
for (int i = 0; i < length; i++){
a[i] = number % 10;
number = number / 10;
index--;
}
Convert the number to String to get the size and use that value to declare the length of the array and then just loop through it, extracting the last number using modulus and then dividing to get rid of the last digit in the integer.

Convert the number into a string, then split the string into an array. After that, convert the characters into integers, like this:
int number = 1234;
List<Integer> digits = ArrayList<Integer>();
String digitsAsString = ("" + number).split("");
for(String digitAsString:digitsAsString){
digits.add(new Integer(Integer.parseInt(digitAsString));
}
Your array of digits is the list of digits

int number = 4567;
String number_string = String.valueOf(number);
char[] chars = number_string.toCharArray();
int[] digits = new int[chars.length];
int i = 0;
for (char c : chars) {
digits[i] = c - '0';
i++;
}

int number = 1234;
String num = Integer.toString(number);
int [] a = new int[num.length()];
for(int i = 0; i < num.length(); i++){
a[i] = Character.getNumericValue(num.charAt(i));
}
First convert number to a String, loop through each char and convert them into an int then storing the value into a.

Importing input file to integer array

This is the content of my Input file:
123
I want to take this input content to a int[] Array
My code:
BufferedReader br = new BufferedReader(new FileReader("yes.txt"));
int[] Array = br.readline().toCharArray(); // Error as it has to be int array
How to solve this:
output:
Array[0]=1
Array[0]=2
Array[0]=3

Simple convert of your string to int array:
private int[] convertStringToIntArray(String str) {
int[] resultArray = new int[str.length()];
for (int i = 0; i < str.length(); i++)
resultArray[i] = Character.getNumericValue(str.charAt(i));
return resultArray;
}

Once you have your string in, it's very simple.
I would recommend using an arraylist for this, as it is easier to add new elements on the fly.
String inputNumbers = "12345";
ArrayList<Integer> numberList = new ArrayList<Integer>();
for(int i = 0; i < inputNumbers.length(); i++) {
numberList.add(Integer.valueOf(numberList.substring(i,i+1)));
}
And there you go! You now have an arraylist called numberList where you can access all of the numbers with numberList.get(0), numberList.get(1), etc.

First read the input as string then convert to int array as below
String inputString = br.readline()
int[] num = new int[inputString.length()];
for (int i = 0; i < inputString.length(); i++){
num[i] = inputString.charAt(i) - '0';
}
Another way is to convert the char array to int array
char[] list = inputString.toCharArray();
int[] num = new int[inputString.length()];
for (int i = 0; i > inputString.length(); i++){
num[i] = list[i];
}
System.out.println(Arrays.toString(num));

Is there a better (faster) way to divide a number to digits?

I wrote this:
void blah(int num)
{
int numOfDigits = Math.log10(num);
int arr[] = new int[numOfDigits + 1];
for(int i = numOfDigits; i>0; i--)
{
arr[i] = num%10;
num = num/10;
}
}
But I thought there must be a more elegant way of doing this. Is there?

If you're happy with Strings, you could do this:
String[] arr = Integer.toString(num).split("(?<=\\d)");
If you want to go from this to int[]:
int[] arrint = new int[arr.length];
for (int i = 0; i < arr.length; i++)
arrint[i] = Integer.parseInt(arr[i]);

Consider converting the integer to a string using Integer.toString and then using string.toCharArray() and converting back. This may or may not be more 'elegant' depending on your worldview.

This might be one way (i am not super proud about this :))
Integer number = 1234567890; // input number
String temp = number.toString(); // convert to string
int [] output = new int[temp.length()];
for (int i=0 ; i< temp.length(); i++) // get character at index i from string
output[i] = temp.charAt(i) - '0'; // convert it to number by removing '0'