MoveNonZeroToTheLeftOfZeroElementsInAnArray

Question: Given an array of integers. Move all non-zero elements to the left of all zero elements.

Source: http://www.careercup.com/question?id=5668212962230272

Solution: 
I have given, two solutions:
1. Using Single Pointer in the array
2. Using Two Pointers in the array

Both have Time Complexity = O(n) and Space Complexity = O(1)

************************************ Using Two Pointers ****************************************************
package MoveNonZeroToTheLeftOfZeroElementsInAnArray;

import java.util.Scanner;

public class UsingTwoPointersInTheArray {
public static void main(String[] args) {
	Scanner in = new Scanner(System.in);
	try{
		System.out.println("Enter the number of elements of the array");
		int n = in.nextInt();
		System.out.println("Enter the elements - zero and non-zero");
		int[] a = new int[n];
		for(int i=0;i<n;i++)
			a[i]=in.nextInt();
		a = moveElements(a);
		printArray(a);
	}
	finally{
		in.close();
	}
}

private static void printArray(int[] a) {
	for(int i=0;i<a.length;i++)
		System.out.print(a[i]+" ");
	System.out.println();
}

private static int[] moveElements(int[] a) {
	int low = 0;
	int high = a.length-1;
	while(low<high){
		if(a[low]==0 && a[high]!=0){
			a=swap(a,low,high);
			high--;
			low++;
			continue;
			}
		if(a[low]==0 && a[high]==0){
			high--;
		}
		if(a[low]!=0 && a[high]==0){
			low++;
			high--;
			}
		if(a[low]!=0 && a[high]!=0){
			low++;
		}
		}
	return a;
	}

private static int[] swap(int[] a, int low, int high) {
	a[low] = a[low]^a[high];
	a[high] = a[low]^a[high];
	a[low] = a[low]^a[high];
	return a;
	}
}
/*
Analysis:
	Time Complexity = O(n)
	Space Complexity = O(1)
*/

************************************ Using Single Pointer ****************************************************
package MoveNonZeroToTheLeftOfZeroElementsInAnArray;

import java.util.Scanner;

public class UsingSinglePointer {
	public static void main(String[] args) {
	Scanner in = new Scanner(System.in);
	try{
	System.out.println("Enter the number of elements in the array");
	int n = in.nextInt();
	System.out.println("Enter the elements of the array - zero and non zero");
	int[] a = new int[n];
	for(int i=0;i<n;i++)
		a[i] = in.nextInt();
	a = moveElements(a);
	printArray(a);
	}
	finally{
		in.close();
		}
	}
	private static int[] moveElements(int[] a) {
		int zeroPoisitionPointer = 0;    // initialize a pointer which points to the element which is 0 in the array.
		                                 // assume initially this pointer is at position 0
		for(int i=0;i<a.length;i++){     // This loop is only for iterating, it doesn't act like a second pointer.
										 // Hence, there is only one pointer in this program(zeroPositionPointer)
			if(a[i]!=0){                 // if the element is not 0, then shift to 0 position
				a[zeroPoisitionPointer++] = a[i]; 
			}
		}
		for(;zeroPoisitionPointer<a.length;zeroPoisitionPointer++) // fill the remaining array elements with 0
			a[zeroPoisitionPointer]=0;
		return a;
	}
	private static void printArray(int[] a) {
		for(int i=0;i<a.length;i++)
			System.out.print(a[i]+" ");
		System.out.println();
	}

}
/*
Analysis:
	Time Complexity = O(n)
	Space Complexity = O(1)
*/