CODEWITHSUNDEEP
javaarrays
Zzzron135
Zzzron135

Reputation: 11

Trying to place a number's digits into an array, but it's reversed

I am attempting to store the digits of a 4-digit number into an array, but it ends up being reversed: 1234 is stored as [4,3,2,1] in the array.

int[] nums = new int[4];

for (int i = 0; i < length; i++) {
    nums[i] = input % 10;
    input /= 10;
    System.out.print(nums[i]);

}

Upvotes: 0

Views: 60

Answers (3)

Filipp Voronov
Filipp Voronov

Reputation: 4197

Another option is to change array index:

nums[nums.length - i - 1] = input % 10;

Upvotes: 0

Kakarot
Kakarot

Reputation: 4252

You just need to traverse the array in a reverse order, since number % 10 will give you the digit in the ten's position.

eg: if number 1234 , number % 10 => 4 which should be stored at the end of your array viz nums[3]

for (int i = nums.length - 1; i >= 0; i--) {
    nums[i] = input % 10;
    input /= 10;
    System.out.print(nums[i]);

}

Upvotes: 2

OneCricketeer
OneCricketeer

Reputation: 191738

input % 10; will return the tens position of the number first, so just loop backwards over the array positions. 

for (int i = nums.length - 1; i >= 0; i--) {

You could also use an ArrayList and check while(input > 0) to do any length number.

Upvotes: 2

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