C Program For Segregating Even And Odd Nodes In A Linked List

Last Updated : 28 Dec, 2022

Given a Linked List of integers, write a function to modify the linked list such that all even numbers appear before all the odd numbers in the modified linked list. Also, keep the order of even and odd numbers the same.

Examples:

Input: 17->15->8->12->10->5->4->1->7->6->NULL Output: 8->12->10->4->6->17->15->5->1->7->NULL  Input: 8->12->10->5->4->1->6->NULL Output: 8->12->10->4->6->5->1->NULL  // If all numbers are even then do not change the list Input: 8->12->10->NULL Output: 8->12->10->NULL  // If all numbers are odd then do not change the list Input: 1->3->5->7->NULL Output: 1->3->5->7->NULL

Method:
The idea is to get pointer to the last node of list. And then traverse the list starting from the head node and move the odd-valued nodes from their current position to end of the list.
Thanks to blunderboy for suggesting this method. Algorithm:

Get pointer to the last node.
Move all the odd nodes to the end.
- Consider all odd nodes before the first even node and move them to end.
- Change the head pointer to point to the first even node.
- Consider all odd nodes after the first even node and move them to the end.

Below is the implementation of the above approach:

// C program to segregate even and  // odd nodes in a Linked List #include <stdio.h> #include <stdlib.h>  // A node of the singly linked list  struct Node {     int data;     struct Node *next; };  void segregateEvenOdd(struct Node **head_ref) {     struct Node *end = *head_ref;     struct Node *prev = NULL;     struct Node *curr = *head_ref;      // Get pointer to the last node      while (end->next != NULL)         end = end->next;      struct Node *new_end = end;      /* Consider all odd nodes before the         first even node and move then after         end */     while (curr->data % 2 != 0 &&             curr != end)     {         new_end->next = curr;         curr = curr->next;         new_end->next->next = NULL;         new_end = new_end->next;     }      // 10->8->17->17->15     /* Do following steps only if there         is any even node */     if (curr->data % 2 == 0)     {         /* Change the head pointer to point             to first even node */         *head_ref = curr;          /* Now current points to the first             even node */         while (curr != end)         {             if ( (curr->data) % 2 == 0 )             {                 prev = curr;                 curr = curr->next;             }             else             {                 /* Break the link between prev                     and current */                 prev->next = curr->next;                  // Make next of curr as NULL                   curr->next = NULL;                  // Move curr to end                  new_end->next = curr;                  // Make curr as new end of list                  new_end = curr;                  /* Update current pointer to next                     of the moved node */                 curr = prev->next;             }         }     }      /* We must have prev set before executing         lines following this statement */     else prev = curr;      /* If there are more than 1 odd nodes and         end of original list is odd then move         this node to end to maintain same order         of odd numbers in modified list */     if (new_end != end &&         (end->data) % 2 != 0)     {         prev->next = end->next;         end->next = NULL;         new_end->next = end;     }     return; }  // UTILITY FUNCTIONS  /* Function to insert a node at     the beginning  */ 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 nodes in  // a given linked list  void printList(struct Node *node) {     while (node!=NULL)     {         printf("%d ", node->data);         node = node->next;     } }  // Driver code int main() {     // Start with the empty list      struct Node* head = NULL;      /* Let us create a sample linked list         as following 0->2->4->6->8->10->11 */     push(&head, 11);     push(&head, 10);     push(&head, 8);     push(&head, 6);     push(&head, 4);     push(&head, 2);     push(&head, 0);      printf("Original Linked list ");     printList(head);      segregateEvenOdd(&head);      printf("Modified Linked list ");     printList(head);      return 0; }

Output

Original Linked list 0 2 4 6 8 10 11 Modified Linked list 0 2 4 6 8 10 11

Time complexity: O(n)
Auxiliary space: O(1) as it is using constant space

Please refer complete article on Segregate even and odd nodes in a Linked List for more details!

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C Program For Segregating Even And Odd Nodes In A Linked List

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