如何在Java中将数组中的数字排序为两个不同的数组？

Question

I have to create a program that takes an array of both even and odd numbers and puts all the even numbers into one array and all the odd numbers into another. I used a for loop to cycle through all the numbers and determine if they are even or odd, but the problem I'm having is that since the numbers in the original array are random, I don't know the size of either the even or the odd array and therefore can't figure out how to assign numbers in the original array to the even/odd arrays without having a bunch of spots left over, or not having enough spots for all the numbers. Any ideas?

Answer 1

Try using an ArrayList. You can use

num % 2 == 0

to see if num is even or odd. If it does == 0 then it is even, else it is odd.

List<Integer> odds = new ArrayList(); 
List<Integer> evens = new ArrayList();

for (int i = 0; i< array.length; i++) {
   if (array[i] % 2 == 0) {
      evens.add(array[i]);
   }
   else {
       odds.add(array[i]);
   }
}

to convert the ArrayLists back to arrays you can do

int[] evn = evens.toArray(new Integer[evens.size()]);

(Note: untested code so there could be a few typos)

EDIT:

If you are not allowed to use ArrayLists then consider the following that just uses Arrays. It's not as efficient as it has to do two passes of the original array

int oddSize = 0; 
int evenSize = 0; 

for (int i = 0; i< array.length; i++) {
   if (array[i] % 2 == 0) {
      evenSize++;
   }
   else {
      oddSize++;
   }
}


Integer[] oddArray = new Integer[oddSize];
Integer[] evenArray = new Integer[evenSize];

int evenIdx = 0;
int oddIdx = 0;

for (int i = 0; i< array.length; i++) {
   if (array[i] % 2 == 0) {
      evenArray[evenIdx++] = array[i];
   }
   else {
      oddArray[oddIdx++] = array[i];
   }
}

Answer 2

You can do it without using arrays or any '%' Just a simple idea

input = new Scanner(System.in);
    int x;
    int y = 0; // Setting Y for 0 so when you add 2 to it always gives even
                // numbers
    int i = 1; // Setting X for 1 so when you add 2 to it always gives odd
                // numbers
    // So for example 0+2=2 / 2+2=4 / 4+2=6 etc..
    System.out.print("Please input a number: ");
    x = input.nextInt();

    for (;;) { // infinite loop so it keeps on adding 2 until the number you
                // input is = to one of y or i
        if (x == y) {
            System.out.print("The number is even ");
            System.exit(0);
        }
        if (x == i) {
            System.out.print("The number is odd ");
            System.exit(0);
        }
        if (x < 0) {
            System.out.print("Invald value");
            System.exit(0);
        }
        y = y + 2;
        i = i + 2;

    }

}

Answer 3

Use a List instead. Then you don't need to declare the sizes in advance, they can grow dynamically.

You can always use the toArray() method on the List afterwards if you really need an array.

Answer 4

The above answers are correct and describe how people would normally implement this. But the description of your problem makes me think this is a class assignment of sorts where dynamic lists are probably unwelcome.

So here's an alternative.

Sort the array to be divided into two parts - of odd and of even numbers. Then count how many odd/even numbers there are and copy the values into two arrays.

Something like this:

static void insertionSort(final int[] arr) {
    int i, j, newValue;
    int oddity;
    for (i = 1; i < arr.length; i++) {
        newValue = arr[i];
        j = i;
        oddity = newValue % 2;
        while (j > 0 && arr[j - 1] % 2 > oddity) {
            arr[j] = arr[j - 1];
            j--;
        }
        arr[j] = newValue;
    }

}

public static void main(final String[] args) {
    final int[] numbers = { 1, 3, 5, 2, 2 };
    insertionSort(numbers);

    int i = 0;
    for (; i < numbers.length; i++) {
        if (numbers[i] % 2 != 0) {
            i--;
            break;
        }
    }

    final int[] evens = new int[i + 1];
    final int[] odds = new int[numbers.length - i - 1];
    if (evens.length != 0) {
        System.arraycopy(numbers, 0, evens, 0, evens.length);
    }
    if (odds.length != 0) {
        System.arraycopy(numbers, i + 1, odds, 0, odds.length);
    }

    for (int j = 0; j < evens.length; j++) {
        System.out.print(evens[j]);
        System.out.print(" ");
    }
    System.out.println();
    for (int j = 0; j < odds.length; j++) {
        System.out.print(odds[j]);
        System.out.print(" ");
    }
}

Answer 5

Iterate through your source array twice. The first time through, count the number of odd and even values. From that, you'll know the size of the two destination arrays. Create them, and take a second pass through your source array, this time copying each value to its appropriate destination array.

Answer 6

I imagine two possibilities, if you can't use Lists, you can iterate twice to count the number of even and odd numbers and then build two arrays with that sizes and iterate again to distribute numbers in each array, but thissolution is slow and ugly.

I imagine another solution, using only one array, the same array that contains all the numbers. You can sort the array, for example set even numbers in the left side and odd numbers in the right side. Then you have one index with the position in the array with the separation ofthese two parts. In the same array, you have two subarrays with the numbers. Use a efficient sort algorithm of course.

Answer 7

Use following Code :

public class ArrayComparing {

    Scanner console= new Scanner(System.in);
    String[] names;
    String[] temp;
    int[] grade;

    public static void main(String[] args) {
        new ArrayComparing().getUserData();
    }

    private void getUserData() {
        names = new String[3];    
        for(int i = 0; i < names.length; i++) {
            System.out.print("Please Enter Student name: ");
            names[i] =console.nextLine();
            temp[i] = names[i];
        }

        grade = new int[3];    
        for(int i =0;i<grade.length;i++) {
            System.out.print("Please Enter Student marks: ");
            grade[i] =console.nextInt();
        }          

        sortArray(names);     
    }

    private void sortArray(String[] arrayToSort) {
        Arrays.sort(arrayToSort);

        getIndex(arrayToSort);
    }

    private void getIndex(String[] sortedArray) {
        for(int x = 0; x < sortedArray.length; x++) {
            for(int y = 0; y < names.length; y++) {
                if(sortedArray[x].equals(temp[y])) {
                    System.out.println(sortedArray[x] + " " + grade[y]);
                }
            }
        }
    }
}