Move all zeros to the front of the linked list

Given a linked list. the task is to move all 0’s to the front of the linked list. The order of all other elements except 0 should be the same after rearrangement.

Examples:

Input : 0 1 0 1 2 0 5 0 4 0
Output :0 0 0 0 0 1 1 2 5 4

Input :1 1 2 3 0 0 0 
Output :0 0 0 1 1 2 3

A simple solution is to store all linked list element in an array. Then move all elements of the array to the beginning. Finally, copy array elements back to the linked list.

An efficient solution is to traverse the linked list from second node. For every node with 0 value, we disconnect it from its current position and move the node to front. 

Implementation:

C++

// CPP program to move all zeros to the end of the linked list.
#include <bits/stdc++.h>
using namespace std;

/* Link list node */
struct Node {
    int data;
    struct Node* next;
};

// Given a reference (pointer to pointer) to the head of a
// list and an int, push a new node on the front of the
// list.
void push(struct Node** head_ref, int new_data)
{
    struct Node* new_node = new Node;
    new_node->data = new_data;
    new_node->next = (*head_ref);
    (*head_ref) = new_node;
}

/* moving zeroes to the beginning in linked list */
void moveZeroes(struct Node** head)
{
    if (*head == NULL)
        return;

    // Traverse the list from second node.
    struct Node *temp = (*head)->next, *prev = *head;
    while (temp != NULL) {

        // If current node is 0, move to beginning of linked
        // list
        if (temp->data == 0) {
            // Disconnect node from its
            // current position
            Node* curr = temp;
            temp = temp->next;
            prev->next = temp;
            // Move to beginning
            curr->next = (*head);
            *head = curr;
        }
        // For non-zero values
        else {
            prev = temp;
            temp = temp->next;
        }
    }
}

// function to displaying nodes
void display(struct Node* head)
{
    while (head != NULL) {
        cout << head->data << "->";
        head = head->next;
    }
    cout << "NULL";
}

/* Driver program to test above function*/
int main()
{
    /* Start with the empty list */
    struct Node* head = NULL;
    /* Use push() to construct below list
    0->0->1->0->1->0->2->0->3->0 */
    push(&head, 0);
    push(&head, 3);
    push(&head, 0);
    push(&head, 2);
    push(&head, 0);
    push(&head, 1);
    push(&head, 0);
    push(&head, 1);
    push(&head, 0);
    push(&head, 0);

    // displaying list before rearrangement
    cout << "Linked list before rearrangement\n";
    display(head);
    /* Check the move_zeroes function */
    moveZeroes(&head);
    // displaying list after rearrangement
    cout << "\n Linked list after rearrangement \n";
    display(head);
    return 0;
}

// This code is contributed by Aditya Kumar (adityakumar129)

C

// C program to move all zeros to the end of the linked list.
#include<stdio.h>
#include<stdlib.h>

/* Link list node */
typedef struct Node {
    int data;
    struct Node* next;
}Node;

// Given a reference (pointer to pointer) to the head of a
// list and an int, push a new node on the front of the
// list.
void push(struct Node** head_ref, int new_data)
{
    struct Node* new_node = (Node *)malloc(sizeof(Node));
    new_node->data = new_data;
    new_node->next = (*head_ref);
    (*head_ref) = new_node;
}

/* moving zeroes to the beginning in linked list */
void moveZeroes(struct Node** head)
{
    if (*head == NULL)
        return;

    // Traverse the list from second node.
    struct Node *temp = (*head)->next, *prev = *head;
    while (temp != NULL) {
        // If current node is 0, move to beginning of linked
        // list
        if (temp->data == 0) {
            // Disconnect node from its current position
            Node* curr = temp;
            temp = temp->next;
            prev->next = temp;
            // Move to beginning
            curr->next = (*head);
            *head = curr;
        }

        // For non-zero values
        else {
            prev = temp;
            temp = temp->next;
        }
    }
}

// function to displaying nodes
void display(struct Node* head)
{
    while (head != NULL) {
        printf("%d->",head->data);
        head = head->next;
    }
    printf("NULL");
}

/* Driver program to test above function*/
int main()
{
    /* Start with the empty list */
    struct Node* head = NULL;
    /* Use push() to construct below list
    0->0->1->0->1->0->2->0->3->0 */
    push(&head, 0);
    push(&head, 3);
    push(&head, 0);
    push(&head, 2);
    push(&head, 0);
    push(&head, 1);
    push(&head, 0);
    push(&head, 1);
    push(&head, 0);
    push(&head, 0);

    // displaying list before rearrangement
    printf("Linked list before rearrangement\n");
    display(head);
    /* Check the move_zeroes function */
    moveZeroes(&head);
    // displaying list after rearrangement
    printf("\nLinked list after rearrangement \n");
    display(head);
    return 0;
}

// This code is contributed by Aditya Kumar (adityakumar129)

Java

// JAVA program to move all zeros
// to the end of the linked list.
class GFG {

    /* Link list node */
    static class Node {
        int data;
        Node next;
    };

    /* Given a reference (pointer to pointer) to
    the head of a list and an int, push a new
    node on the front of the list. */
    static Node push(Node head_ref, int new_data)
    {
        Node new_node = new Node();
        new_node.data = new_data;
        new_node.next = head_ref;
        head_ref = new_node;
        return new_node;
    }

    /* moving zeroes to the beginning in linked list */
    static Node moveZeroes(Node head)
    {
        if (head == null)
            return null;

        // Traverse the list from second node.
        Node temp = (head).next, prev = head;
        while (temp != null) {

            // If current node is 0, move to
            // beginning of linked list
            if (temp.data == 0) {

                // Disconnect node from its
                // current position
                Node curr = temp;
                temp = temp.next;
                prev.next = temp;

                // Move to beginning
                curr.next = (head);
                head = curr;
            }

            // For non-zero values
            else {
                prev = temp;
                temp = temp.next;
            }
        }
        return head;
    }

    // function to displaying nodes
    static void display(Node head)
    {
        while (head != null) {
            System.out.print(head.data + "->");
            head = head.next;
        }
        System.out.print("null");
    }

    /* Driver code*/
    public static void main(String[] args)
    {

        /* Start with the empty list */
        Node head = null;

        /* Use push() to construct below list
        0.0.1.0.1.0.2.0.3.0 */
        head = push(head, 0);
        head = push(head, 3);
        head = push(head, 0);
        head = push(head, 2);
        head = push(head, 0);
        head = push(head, 1);
        head = push(head, 0);
        head = push(head, 1);
        head = push(head, 0);
        head = push(head, 0);

        // displaying list before rearrangement
        System.out.print(
            "Linked list before rearrangement\n");
        display(head);

        /* Check the move_zeroes function */
        head = moveZeroes(head);

        // displaying list after rearrangement
        System.out.print(
            "\n Linked list after rearrangement \n");
        display(head);
    }
}

// This code is contributed by Aditya Kumar (adityakumar129)

Python3

# Python3 program to move all zeros
# to the end of the linked list.

''' Link list node '''
class Node:
    def __init__(self, data):
        self.data = data
        self.next = None

''' Given a reference (pointer to pointer) to
the head of a list and an int, push a new
node on the front of the list. '''
def push(head_ref, new_data):
  new_node = Node(new_data)
  new_node.next = (head_ref);
  (head_ref) = new_node;
  return head_ref

''' moving zeroes to the beginning in linked list '''
def moveZeroes(head):

  if (head == None):
    return;

  # Traverse the list from second node.
  temp = head.next
  prev = head;
  while (temp != None):

    # If current node is 0, move to
    # beginning of linked list
    if (temp.data == 0):

      # Disconnect node from its
      # current position
      curr = temp;
      temp = temp.next;
      prev.next = temp;

      # Move to beginning
      curr.next = (head);
      head = curr;

    # For non-zero values
    else:
      prev = temp;
      temp = temp.next;
  return head

# function to displaying nodes
def display(head):

  while (head != None):
      print(head.data, end = '->')
      head = head.next;

  print('NULL', end = '')

''' Driver program to test above function'''
if __name__=='__main__':

  ''' Start with the empty list '''
  head = None;

  ''' Use push() to construct below list
  0.0.1.0.1.0.2.0.3.0 '''
  head = push(head, 0);
  head = push(head, 3);
  head = push(head, 0);
  head = push(head, 2);
  head = push(head, 0);
  head = push(head, 1);
  head = push(head, 0);
  head = push(head, 1);
  head = push(head, 0);
  head = push(head, 0);

  # displaying list before rearrangement
  print("Linked list before rearrangement");
  display(head);

  ''' Check the move_zeroes function '''
  head = moveZeroes(head);

  # displaying list after rearrangement
  print("\n\nLinked list after rearrangement ");
  display(head);

# This code is contributed by rutvik_56

C#

// C# program to move all zeros
// to the end of the linked list.
using System;

class GFG
{

    /* Link list node */
    class Node
    {
        public int data;
        public Node next;
    };

    /* Given a reference (pointer to pointer) to
    the head of a list and an int, push a new
    node on the front of the list. */
    static Node push(Node head_ref, int new_data)
    {
        Node new_node = new Node();
        new_node.data = new_data;
        new_node.next = head_ref;
        head_ref = new_node;
        return new_node;
    }

    /* moving zeroes to the beginning in linked list */
    static Node moveZeroes(Node head)
    {
        if (head == null)
            return null;

        // Traverse the list from second node.
        Node temp = (head).next, prev = head;
        while (temp != null)
        {

            // If current node is 0, move to
            // beginning of linked list
            if (temp.data == 0)
            {

                // Disconnect node from its
                // current position
                Node curr = temp;
                temp = temp.next;
                prev.next = temp;

                // Move to beginning
                curr.next = (head);
                head = curr;
            }

            // For non-zero values
            else
            {
                prev = temp;
                temp = temp.next;
            }
        }
        return head;
    }

    // function to displaying nodes
    static void display(Node head)
    {
        while (head != null)
        {
            Console.Write(head.data + "->");
            head = head.next;
        }
        Console.Write("null");
    }

    // Driver code
    public static void Main(String[] args)
    {

        /* Start with the empty list */
        Node head = null;

        /* Use push() to construct below list
        0->0->1->0->1->0->2->0->3->0 */
        head = push(head, 0);
        head = push(head, 3);
        head = push(head, 0);
        head = push(head, 2);
        head = push(head, 0);
        head = push(head, 1);
        head = push(head, 0);
        head = push(head, 1);
        head = push(head, 0);
        head = push(head, 0);

        // displaying list before rearrangement
        Console.Write("Linked list before rearrangement\n");
        display(head);

        /* Check the move_zeroes function */
        head = moveZeroes(head);

        // displaying list after rearrangement
        Console.Write("\n Linked list after rearrangement \n");
        display(head);
    }
}

// This code is contributed by Rajput-Ji

Javascript

<script>

// JavaScript program to move all zeros
// to the end of the linked list.

/* Link list node */
class Node
{
    constructor()
    {
        this.data = 0;
        this.next = null;
    }
};

/* Given a reference (pointer to pointer) to
the head of a list and an int, push a new
node on the front of the list. */
function push(head_ref, new_data)
{
    var new_node = new Node();
    new_node.data = new_data;
    new_node.next = head_ref;
    head_ref = new_node;
    return new_node;
}

/* moving zeroes to the beginning in linked list */
function moveZeroes(head)
{
    if (head == null)
        return null;

    // Traverse the list from second node.
    var temp = (head).next, prev = head;
    while (temp != null)
    {

        // If current node is 0, move to
        // beginning of linked list
        if (temp.data == 0)
        {

            // Disconnect node from its
            // current position
            var curr = temp;
            temp = temp.next;
            prev.next = temp;

            // Move to beginning
            curr.next = (head);
            head = curr;
        }

        // For non-zero values
        else
        {
            prev = temp;
            temp = temp.next;
        }
    }
    return head;
}

// function to displaying nodes
function display(head)
{
    while (head != null)
    {
        document.write(head.data + "->");
        head = head.next;
    }
    document.write("null");
}

// Driver code
/* Start with the empty list */
var head = null;

/* Use push() to construct below list
0->0->1->0->1->0->2->0->3->0 */
head = push(head, 0);
head = push(head, 3);
head = push(head, 0);
head = push(head, 2);
head = push(head, 0);
head = push(head, 1);
head = push(head, 0);
head = push(head, 1);
head = push(head, 0);
head = push(head, 0);

// displaying list before rearrangement
document.write("Linked list before rearrangement<br>");
display(head);

/* Check the move_zeroes function */
head = moveZeroes(head);

// displaying list after rearrangement
document.write("<br> Linked list after rearrangement <br>");
display(head);

</script>

Output

Linked list before rearrangement
0->0->1->0->1->0->2->0->3->0->NULL
 Linked list after rearrangement 
0->0->0->0->0->0->1->1->2->3->NULL

Time Complexity: O(n), where n is the size of the given list.

Auxiliary Space: O(1) because it is using constant space