将整数数组的前半部分和奇数移动到后半部分

Question

I had an interview question which i could not solve.

Write method (not a program) in Java Programming Language that will move all even numbers on the first half and odd numbers to the second half in an integer array.

Eg Input = {3,8,12,5,9,21,6,10}; Output = {12,8,6,10,3,5,9,21}.

The method should take integer array as parameter and move items in the same array (do not create another array). The numbers may be in different order than original array. This is algorithm test, so try to give as efficient algorithm as you can (possibly linear O(n) algorithm). Avoid using built in functions/API. *

Also some basic intro to what is data structure efficiency

Answer 1

Keep two indices: one to the first odd number and one to the last even number. Swap such numbers and update indices.

Answer 2

(With a lot of help from @manu-fatto's suggestion) I believe this would do it:

private static int[] OddSort(int[] items)
{
    int oddPos, nextEvenPos;
    for (nextEvenPos = 0; 
         nextEvenPos < items.Length && items[nextEvenPos] % 2 == 0;
         nextEvenPos++) { }
    // nextEvenPos is now positioned at the first odd number in the array, 
    // i.e. it is the next place an even number will be placed

    // We already know that items[nextEvenPos] is odd (from the condition of the 
    // first loop), so we'll start looking for even numbers at nextEvenPos + 1
    for (oddPos = nextEvenPos + 1; oddPos < items.Length; oddPos++)
    {
        // If we find an even number
        if (items[oddPos] % 2 == 0)
        {
            // Swap the values
            int temp = items[nextEvenPos];
            items[nextEvenPos] = items[oddPos];
            items[oddPos] = temp;
            // And increment the location for the next even number
            nextEvenPos++;
        }
    }

    return items;
}

This algorithm traverses the list exactly 1 time (inspects each element exactly once), so the efficiency is O(n).

Answer 3

// to do this in one for loop

public static void evenodd(int[] integer) {

    int i = 0, temp = 0;
    int j = integer.length - 1;

    while (j >= i) {
        // swap if found odd even combo at i and j
        if (integer[i] % 2 != 0 && integer[j] % 2 == 0) {
            temp = integer[i];
            integer[i] = integer[j];
            integer[j] = temp;
            i++;
            j--;

        } else {
            if (integer[i] % 2 == 0) {
                i++;
            }
            if (integer[j] % 2 == 1) {
                j--;
            }

        }

    }
} 

Answer 4

@JLRishe,

Your algorithm doesn't maintain the order. For a simple example, say {1,5,2}, you will change the array to {2,5,1}. I could not comment below your post as I am a new user and lack reputations.

Answer 5

public static void sorted(int [] integer) {

int i, j , temp;

for (i = 0;  i < integer.length;  i++) {

     if (integer[i] % 2 == 0) {
         for (j = i;  j < integer.length;  j++) {
              if (integer[j] % 2 == 1) {
                  temp = y[i];
                  y[i] = y[j];
                  y[j] = temp;
              }
          }
      }
      System.out.println(integer[i]);
}

public static void main(String args[]) {

       sorted(new int[]{1, 2,7, 9, 4}); 



}

}

The answer is 1, 7, 9, 2, 4.

Answer 6

Could it be that you were asked to implement a very basic version of the BubbleSort where the sort value of element e, where e = arr[i], = e%2==1 ? 1 : -1 ? Regards Leon

Answer 7

class Demo
{
public void sortArray(int[] a)
{
int len=a.length;
int j=len-1;
for(int i=0;i<len/2+1;i++)
{
if(a[i]%2!=0)
{
while(a[j]%2!=0 && j>(len/2)-1)
j--;
if(j<=(len/2)-1)
break;
a[i]=a[i]+a[j];
a[j]=a[i]-a[j];
a[i]=a[i]-a[j];
}
}
for(int i=0;i<len;i++)
System.out.println(a[i]);
}

public static void main(String s[])
{
int a[]=new int[10];
System.out.println("Enter 10 numbers");
java.util.Scanner sc=new java.util.Scanner(System.in);
for(int i=0;i<10;i++)
{
a[i]=sc.nextInt();
}
new Demo().sortArray(a);
}
}

Answer 8

private static void rearrange(int[] a) {
    int i,j,temp;
    for(i = 0, j = a.length - 1; i < j ;i++,j--) {
        while(a[i]%2 == 0 && i != a.length - 1) {
            i++;
        }
        while(a[j]%2 == 1 && j != 0) {
            j--;
        }
        if(i>j)
            break;
        else {
            temp = a[i];
            a[i] = a[j];
            a[j] = temp;
        }
    }       
}

Answer 9

public void sortEvenOddIntegerArray(int[] intArray){
    boolean loopRequired = false;
    do{
        loopRequired = false;
        for(int i = 0;i<intArray.length-1;i++){

            if(intArray[i] % 2 != 0 && intArray[i+1] % 2 == 0){

                int temp = intArray[i];
                intArray[i] = intArray[i+1];
                intArray[i+1] = temp;
                loopRequired = true;
            }
        }
    }while(loopRequired);
}

Answer 10

You can do this with a single loop by moving odd items to the end of the array when you find them.

static void EvensToLeft(int[] items) {
    int end = items.length;
    for (int i = 0; i < end; i++) {
        if (items[i] % 2) {
            int t = items[i];
            items[i--] = items[--end];
            items[end] = t;
        }
    }
}

Given an input array of length n the inner loop executes exactly n times, and computes the parity of each array element exactly once.

Answer 11

Use two counters i=0 and j=a.length-1 and keep swapping even and odd elements that are in the wrong place.

  public int[] evenOddSort(int[] a) {
        int i = 0;
        int j = a.length - 1;
        int temp;
        while (i < j) {
            if (a[i] % 2 == 0) {
                i++;
            } else if (a[j] % 2 != 0) {
                j--;
            } else {
                temp = a[i];
                a[i] = a[j];
                a[j] = temp;
                i++;
                j--;
            }
        }
        return a;
    }

Answer 12

public class SeperatOddAndEvenInList {

public static int[] seperatOddAndEvnNos(int[] listOfNumbers) {
    int oddNumPointer = 0;
    int evenNumPointer = listOfNumbers.length - 1;
    while(oddNumPointer <= evenNumPointer) {                
            if(listOfNumbers[oddNumPointer] % 2 == 0) { //even number, swap to front of last known even number
                int temp;
                temp = listOfNumbers[oddNumPointer];
                listOfNumbers[oddNumPointer] = listOfNumbers[evenNumPointer];
                listOfNumbers[evenNumPointer] = temp;
                evenNumPointer--;
            }
            else {  //odd number, go ahead... capture next element
                oddNumPointer++;
            }


    }
    return listOfNumbers;
}


public static void main(String[] args) {
    // TODO Auto-generated method stub
    int []arr = {3, 8, 12, 5, 9, 21, 6, 10};
    int[] seperatedArray = seperatOddAndEvnNos(arr);
    for (int i : seperatedArray) {
        System.out.println(i);
    }

}

}

Answer 13

efficiency is O(log n).

public class TestProg {
public static void main(String[] args) {
    int[] input = { 32, 54, 35, 18, 23, 17, 2 };
    int front = 0;
    int mid = input.length - 1;
    for (int start = 0; start < input.length; start++) {
    //if current element is odd
        if (start < mid && input[start] % 2 == 1) {
    //swapping element is also odd?
            if (input[mid] % 2 == 1) {
                mid--;
                start--;
            } 
    //swapping element is not odd then swap
     else {
                int tmp = input[mid];
                input[mid] = input[start];
                input[start] = tmp;
                mid--;
            }
        }

    }
    for (int x : input)
        System.out.print(x + " ");
}

}

Answer 14

public class ArraysSortEvensFirst {

public static void main(String[] args) {
    int[] arr = generateTestData();
    System.out.println(Arrays.toString(arr));

    ArraysSortEvensFirst test = new ArraysSortEvensFirst();
    test.sortEvensFirst(arr);

}

private static int[] generateTestData() {
    int[] arr = {1,3,5,6,9,2,4,5,7};
    return arr;
}

public int[] sortEvensFirst(int[] arr) {
    int end = arr.length;

    int last = arr.length-1;
    for(int i=0; i < arr.length; i++) {
        // find odd elements, then move to even slots
        if(arr[i]%2 > 0) {
            int k = findEven(last, arr);
            if(k > i) swap(arr, i, k);
            last = k;
        }
    }

    System.out.println(Arrays.toString(arr));
    return arr;
}

public int findEven(int last, int[] arr) {
    for(int k = last; k > 0; k--) {
        if(arr[k]%2 == 0) {
            return k;
        }
    }
    return -1; // not found;
}

public void swap(int[] arr, int x, int y) {
    int temp = arr[x];
    arr[x] = arr[y];
    arr[y] = temp;
}

}

Output: [1, 3, 5, 6, 9, 2, 4, 5, 7] [4, 2, 6, 5, 9, 3, 1, 5, 7]