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
// 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 codeint 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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