shift numbers in an array

Question

I want create a method in java that will take a single dimensional array and shift all the non-zero numbers to the left of the array. I've tried some ways but nothing is working. Can somebody help me? Examples are given below for your reference:

int[] x = {0,8,0,7,0,0,2};
answer = {8,7,2,0,0,0,0}

int[] y = {5,0,0,0,1,0,3};
answer = {5,1,3,0,0,0,0}

That's what I tried:

import java.util.Arrays;

public class Board
    {

    public static final int EMPTY = 0;

    public static boolean alterOneLine(int[] x)
    {
        int[] temp = new int[x.length];
        for(int i=0;i<x.length;i++)
        {
            if (x[i]!=EMPTY)
                temp[i]=x[i];
        }
        if(Arrays.equals(x,temp))
            return false;
        else
            return true;
    }
}

Your code doesn't require a shift of the array because you're implementing a method that returns true if a given array is shifted. Your code is essentially just creating a copy of the original array and does no shift. You have to keep an additional counter for temp that you only increment when assigning a new value, e.g. temp[tmpCounter++] = x[i] — Lothar
– Lothar, Commented Feb 6, 2018 at 21:44
Stop vandalizing your post. If you want to ask another question, post it separately. — shmosel
– shmosel, Commented Feb 6, 2018 at 23:01

shmosel · Accepted Answer · 2018-02-06 22:04:22Z

5

Just iterate and move the non-zero numbers to the beginning:

for (int i = 0, dest = 0; i < arr.length; i++) {
    if (arr[i] != 0) {
        int val = arr[i];
        arr[i] = 0;
        arr[dest++] = val;
    }
}

Or you can make a copy and rely on the array initialization defaults for the zeros:

int[] copy = new int[arr.length];
for (int i = 0, dest = 0; i < arr.length; i++) {
    if (arr[i] != 0) {
        copy[dest++] = arr[i];
    }
}

Alternatively, you can use a boxed stream with a custom sort:

int[] copy = Arrays.stream(arr)
        .boxed()
        .sorted(Comparator.comparing(i -> i == 0))
        .mapToInt(Integer::intValue)
        .toArray();

Or you can use Guava's Ints.asList() and sort the array in-place:

Ints.asList(arr).sort(Comparator.comparing(i -> i == 0));

edited Feb 6, 2018 at 22:04

answered Feb 6, 2018 at 21:47

shmosel

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1 Comment

zlakad Over a year ago

I'm sorry for edit - I totally forgot about initializing primitive and arrays! Up voting.

Lothar · Accepted Answer · 2018-02-06 21:56:54Z

As I said in my comment, your method doesn't require shifting the array because the method only checks if a given is already shifted (and returns false in that case. But for the sake of answering, here is the corrected version:

import java.util.Arrays;

public class Board {

    public static final int EMPTY = 0;

    public static boolean alterOneLine(int[] x) {
        int[] temp = new int[x.length];
        int tempCounter = 0;
        for (int i = 0; i < x.length; i++) {
            if (x[i] != EMPTY)
                temp[tempCounter++] = x[i];
        }
        if (Arrays.equals(x, temp))
            return false;
        else
            return true;
    }

    public static boolean alterOneLine2(int[] x) {
        boolean zeroSeen = false;
        for (int i = 0; i < x.length; i++) {
            if (x[i] != 0 && zeroSeen) {
                return true;
            }
            if (x[i] == 0) {
                zeroSeen = true;
            }
        }
        return false;
    }

    public final static void main(String[] args) {
        System.out.println(alterOneLine(new int[]{0,8,0,7,0,0,2}));
        System.out.println(alterOneLine(new int[]{8,7,2,0,0,0,0}));
        System.out.println(alterOneLine(new int[]{5,0,0,0,1,0,3}));
        System.out.println(alterOneLine(new int[]{5,1,3,0,0,0,0}));
        System.out.println(alterOneLine2(new int[]{0,8,0,7,0,0,2}));
        System.out.println(alterOneLine2(new int[]{8,7,2,0,0,0,0}));
        System.out.println(alterOneLine2(new int[]{5,0,0,0,1,0,3}));
        System.out.println(alterOneLine2(new int[]{5,1,3,0,0,0,0}));
    }
}

Your error was using the same index when assiging a value to temp instead of an individual counter that is only increased if you shift a value to the left.

I also added a second method with the same functionality but without creating a temporary array and supsequent call of Arrays#equals (two iterations over the whole array instead of one with my new method).

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