Skip to content
Related Articles

Related Articles

C++ Program For Partitioning A Linked List Around A Given Value And Keeping The Original Order

View Discussion
Improve Article
Save Article
  • Last Updated : 22 Jun, 2022

Given a linked list and a value x, partition it such that all nodes less than x come first, then all nodes with a value equal to x, and finally nodes with a value greater than or equal to x. The original relative order of the nodes in each of the three partitions should be preserved. The partition must work in place.
Examples:

Input: 1->4->3->2->5->2->3, 
        x = 3
Output: 1->2->2->3->3->4->5

Input: 1->4->2->10 
        x = 3
Output: 1->2->4->10

Input: 10->4->20->10->3 
        x = 3
Output: 3->10->4->20->10 

To solve this problem we can use partition method of Quick Sort but this would not preserve the original relative order of the nodes in each of the two partitions.
Below is the algorithm to solve this problem :

  1. Initialize first and last nodes of below three linked lists as NULL.
    • Linked list of values smaller than x.
    • Linked list of values equal to x.
    • Linked list of values greater than x.
  2. Now iterate through the original linked list. If a node’s value is less than x then append it at the end of the smaller list. If the value is equal to x, then at the end of the equal list. And if a value is greater, then at the end of the greater list.
  3. Now concatenate three lists.

Below is the implementation of the above idea. 

C++




// C++ program to partition a linked list
// around a given value.
#include<bits/stdc++.h>
using namespace std;
 
// Link list Node
struct Node
{
    int data;
    struct Node* next;
};
 
// A utility function to create
// a new node
Node *newNode(int data)
{
    struct Node* new_node = new Node;
    new_node->data  = data;
    new_node->next = NULL;
    return new_node;
}
 
// Function to make two separate lists
// and return head after concatenating
struct Node *partition(struct Node *head,
                       int x)
{
    /* Let us initialize first and last
       nodes of three linked lists
       1) Linked list of values smaller
          than x.
       2) Linked list of values equal
          to x.
       3) Linked list of values greater
          than x.*/
    struct Node *smallerHead = NULL,
                *smallerLast = NULL;
    struct Node *greaterLast = NULL,
                *greaterHead = NULL;
    struct Node *equalHead = NULL,
                *equalLast = NULL;
 
    // Now iterate original list and
    // connect nodes of appropriate
    // linked lists.
    while (head != NULL)
    {
        // If current node is equal to x,
        // append it to the list of x values
        if (head->data == x)
        {
            if (equalHead == NULL)
                equalHead = equalLast = head;
            else
            {
                equalLast->next = head;
                equalLast = equalLast->next;
            }
        }
 
        // If current node is less than X,
        // append it to the list of smaller
        // values
        else if (head->data < x)
        {
            if (smallerHead == NULL)
                smallerLast = smallerHead = head;
            else
            {
                smallerLast->next = head;
                smallerLast = head;
            }
        }
        
        // Append to the list of greater values
        else
        {
            if (greaterHead == NULL)
                greaterLast = greaterHead = head;
            else
            {
                greaterLast->next  = head;
                greaterLast = head;
            }
        }
 
        head = head->next;
    }
 
    // Fix end of greater linked list to
    // NULL if this list has some nodes
    if (greaterLast != NULL)
        greaterLast->next = NULL;
 
    // Connect three lists
 
    // If smaller list is empty
    if (smallerHead == NULL)
    {
        if (equalHead == NULL)
            return greaterHead;
        equalLast->next = greaterHead;
        return equalHead;
    }
 
    // If smaller list is not empty
    // and equal list is empty
    if (equalHead == NULL)
    {
        smallerLast->next = greaterHead;
        return smallerHead;
    }
 
    // If both smaller and equal list
    // are non-empty
    smallerLast->next = equalHead;
    equalLast->next = greaterHead;
    return  smallerHead;
}
 
// Function to print linked list
void printList(struct Node *head)
{
    struct Node *temp = head;
    while (temp != NULL)
    {
        printf("%d  ", temp->data);
        temp = temp->next;
    }
}
 
// Driver code
int main()
{
    // Start with the empty list
    struct Node* head = newNode(10);
    head->next = newNode(4);
    head->next->next = newNode(5);
    head->next->next->next = newNode(30);
    head->next->next->next->next =
    newNode(2);
    head->next->next->next->next->next =
    newNode(50);
 
    int x = 3;
    head = partition(head, x);
    printList(head);
    return 0;
}


Output:  

2 10 4 5 30 50

Time Complexity: O(n) where n is the size of the given linked list.
Auxiliary Space: O(1), no extra space is required, so it is a constant.

Please refer complete article on Partitioning a linked list around a given value and keeping the original order for more details!


My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!