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C Program For Sorting A Linked List Of 0s, 1s And 2s

  • Last Updated : 22 Dec, 2021

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

Input: 1 -> 1 -> 2 -> 0 -> 2 -> 0 -> 1 -> NULL
Output: 0 -> 0 -> 1 -> 1 -> 1 -> 2 -> 2 -> NULL

Input: 1 -> 1 -> 2 -> 1 -> 0 -> NULL 
Output: 0 -> 1 -> 1 -> 1 -> 2 -> NULL

Source: Microsoft Interview | Set 1

Following steps can be used to sort the given linked list.

  • Traverse the list and count the number of 0s, 1s, and 2s. Let the counts be n1, n2, and n3 respectively.
  • Traverse the list again, fill the first n1 nodes with 0, then n2 nodes with 1, and finally n3 nodes with 2.

Below image is a dry run of the above approach:

Below is the implementation of the above approach:

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)
{
    // Initialize count of '0', '1' 
    // and '2' as 0
    int count[3] = {0, 0, 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");
}
  
// Driver code
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 Sorting");
    printList(head);
  
    sortList(head);
  
    printf(
    "Linked List After Sorting");
    printList(head);
  
    return 0;
}

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 the number of nodes in the linked list. 
Auxiliary Space: O(1)

Please refer complete article on Sort a linked list of 0s, 1s and 2s for more details!


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