Sort a linked list of 0s, 1s and 2s

1.7

Given a linked list of 0s, 1s and 2s, sort it.

Source: Microsoft Interview | Set 1

Following steps can be used to sort the given linked list.
1) Traverse the list and count the number of 0s, 1s and 2s. Let the counts be n1, n2 and n3 respectively.
2) Traverse the list again, fill the first n1 nodes with 0, then n2 nodes with 1 and finally n3 nodes with 2.

C/C++

// C Program to sort a linked list 0s, 1s or 2s
#include<stdio.h>
#include<stdlib.h>

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

// Function to sort a linked list of 0s, 1s and 2s
void sortList(struct Node *head)
{
    int count[3] = {0, 0, 0};  // Initialize count of '0', '1' and '2' as 0
    struct Node *ptr = head;

    /* count total number of '0', '1' and '2'
     * count[0] will store total number of '0's
     * count[1] will store total number of '1's
     * count[2] will store total number of '2's  */
    while (ptr != NULL)
    {
        count[ptr->data] += 1;
        ptr = ptr->next;
    }

    int i = 0;
    ptr = head;

    /* Let say count[0] = n1, count[1] = n2 and count[2] = n3
     * now start traversing list from head node,
     * 1) fill the list with 0, till n1 > 0
     * 2) fill the list with 1, till n2 > 0
     * 3) fill the list with 2, till n3 > 0  */
    while (ptr != NULL)
    {
        if (count[i] == 0)
            ++i;
        else
        {
            ptr->data = i;
            --count[i];
            ptr = ptr->next;
        }
    }
}

/* Function to push a node */
void push (struct Node** head_ref, int new_data)
{
    /* allocate node */
    struct Node* new_node =
        (struct Node*) malloc(sizeof(struct Node));

    /* put in the data  */
    new_node->data  = new_data;

    /* link the old list off the new node */
    new_node->next = (*head_ref);

    /* move the head to point to the new node */
    (*head_ref)    = new_node;
}

/* Function to print linked list */
void printList(struct Node *node)
{
    while (node != NULL)
    {
        printf("%d  ", node->data);
        node = node->next;
    }
    printf("n");
}

/* Drier program to test above function*/
int main(void)
{
    struct Node *head = NULL;
    push(&head, 0);
    push(&head, 1);
    push(&head, 0);
    push(&head, 2);
    push(&head, 1);
    push(&head, 1);
    push(&head, 2);
    push(&head, 1);
    push(&head, 2);

    printf("Linked List Before Sortingn");
    printList(head);

    sortList(head);

    printf("Linked List After Sortingn");
    printList(head);

    return 0;
}

Java

// Java program to sort a linked list of 0, 1 and 2
class LinkedList
{
    Node head;  // head of list
 
    /* Linked list Node*/
    class Node
    {
        int data;
        Node next;
        Node(int d) {data = d; next = null; }
    }

    void sortList()
    {
       // initialise count of 0 1 and 2 as 0
       int count[] = {0, 0, 0}; 
       
       Node ptr = head;
       
       /* count total number of '0', '1' and '2'
        * count[0] will store total number of '0's
        * count[1] will store total number of '1's
        * count[2] will store total number of '2's  */
       while (ptr != null) 
       {
            count[ptr.data]++;
            ptr = ptr.next;
       }

       int i = 0;
       ptr = head;

       /* Let say count[0] = n1, count[1] = n2 and count[2] = n3
        * now start traversing list from head node,
        * 1) fill the list with 0, till n1 > 0
        * 2) fill the list with 1, till n2 > 0
        * 3) fill the list with 2, till n3 > 0  */
        while (ptr != null) 
        {
            if (count[i] == 0)
                i++;
            else 
            {
               ptr.data= i;
               --count[i];
               ptr = ptr.next;
            }
         }
    }                       

                   
    /* Utility functions */

    /* Inserts a new Node at front of the list. */
    public void push(int new_data)
    {
        /* 1 & 2: Allocate the Node &
                  Put in the data*/
        Node new_node = new Node(new_data);
 
        /* 3. Make next of new Node as head */
        new_node.next = head;
 
        /* 4. Move the head to point to new Node */
        head = new_node;
    }

    /* Function to print linked list */
    void printList()
    {
        Node temp = head;
        while (temp != null)
        {
           System.out.print(temp.data+" ");
           temp = temp.next;
        }  
        System.out.println();
    }

     /* Drier program to test above functions */
    public static void main(String args[])
    {
        LinkedList llist = new LinkedList();
        
        /* Constructed Linked List is 1->2->3->4->5->6->7->
           8->8->9->null */
        llist.push(0);
        llist.push(1);
        llist.push(0);
        llist.push(2);
        llist.push(1);
        llist.push(1);
        llist.push(2);
        llist.push(1);
        llist.push(2);
        
        System.out.println("Linked List before sorting");
        llist.printList();
        
        llist.sortList();

        System.out.println("Linked List after sorting");
        llist.printList();
    }
} 
/* This code is contributed by Rajat Mishra */

Python


# Python program to sort a linked list of 0, 1 and 2
class LinkedList(object):
    def __init__(self):

         # head of list
         self.head = None

    # Linked list Node
    class Node(object):
        def __init__(self, d):
            self.data = d
            self.next = None

    def sortList(self):

        # initialise count of 0 1 and 2 as 0
        count = [0, 0, 0]

        ptr = self.head

        # count total number of '0', '1' and '2'
        # * count[0] will store total number of '0's
        # * count[1] will store total number of '1's
        # * count[2] will store total number of '2's  
        while ptr != None:
            count[ptr.data]+=1
            ptr = ptr.next

        i = 0
        ptr = self.head

        # Let say count[0] = n1, count[1] = n2 and count[2] = n3
        # * now start traversing list from head node,
        # * 1) fill the list with 0, till n1 > 0
        # * 2) fill the list with 1, till n2 > 0
        # * 3) fill the list with 2, till n3 > 0  
        while ptr != None:
            if count[i] == 0:
                i+=1
            else:
                ptr.data = i
                count[i]-=1
                ptr = ptr.next


    # Utility functions
    # Inserts a new Node at front of the list.
    def push(self, new_data):

        # 1 & 2: Allocate the Node &
        # Put in the data
        new_node = self.Node(new_data)

        # 3. Make next of new Node as head
        new_node.next = self.head

        # 4. Move the head to point to new Node
        self.head = new_node

    # Function to print linked list
    def printList(self):
        temp = self.head
        while temp != None:
            print str(temp.data),
            temp = temp.next
        print ''

# Drier program to test above functions
llist = LinkedList()
llist.push(0)
llist.push(1)
llist.push(0)
llist.push(2)
llist.push(1)
llist.push(1)
llist.push(2)
llist.push(1)
llist.push(2)

print "Linked List before sorting"
llist.printList()

llist.sortList()

print "Linked List after sorting"
llist.printList()

# This code is contributed by BHAVYA JAIN


Output:
Linked List Before Sorting
2  1  2  1  1  2  0  1  0
Linked List After Sorting
0  0  1  1  1  1  2  2  2

Time Complexity: O(n) where n is number of nodes in linked list.
Auxiliary Space: O(1)


Sort a linked list of 0s, 1s and 2s by changing links

This article is compiled by Narendra Kangralkar. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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