C++ Program For Segregating Even And Odd Nodes In A Linked List
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 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 1:
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.
C++
// C++ program to segregate even and// odd nodes in a Linked List#include <bits/stdc++.h>using namespace std;// A node of the singly linked listclass Node{ public: int data; Node *next;};void segregateEvenOdd(Node **head_ref){ Node *end = *head_ref; Node *prev = NULL; Node *curr = *head_ref; // Get pointer to the last node while (end->next != NULL) end = end->next; 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(Node** head_ref, int new_data){ // Allocate node Node* new_node = new 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 listvoid printList(Node *node){ while (node != NULL) { cout << node->data <<" "; node = node->next; }}// Driver codeint main(){ // Start with the empty list 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); cout << "Original Linked list "; printList(head); segregateEvenOdd(&head); cout << "Modified Linked list "; printList(head); return 0;}// This code is contributed by rathbhupendra |
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)
Method 2:
The idea is to split the linked list into two: one containing all even nodes and other containing all odd nodes. And finally, attach the odd node linked list after the even node linked list.
To split the Linked List, traverse the original Linked List and move all odd nodes to a separate Linked List of all odd nodes. At the end of the loop, the original list will have all the even nodes and the odd node list will have all the odd nodes. To keep the ordering of all nodes the same, we must insert all the odd nodes at the end of the odd node list. And to do that in constant time, we must keep track of the last pointer in the odd node list.
C++
// CPP program to segregate even and// odd nodes in a Linked List#include <stdio.h>#include <stdlib.h>// A node of the singly linked liststruct Node{ int data; struct Node *next;};// Function to segregate even and// odd nodes.void segregateEvenOdd(struct Node **head_ref){ // Starting node of list having // even values. Node *evenStart = NULL; // Ending node of even values list. Node *evenEnd = NULL; // Starting node of odd values list. Node *oddStart = NULL; // Ending node of odd values list. Node *oddEnd = NULL; // Node to traverse the list. Node *currNode = *head_ref; while(currNode != NULL) { int val = currNode -> data; // If current value is even, add // it to even values list. if(val % 2 == 0) { if(evenStart == NULL) { evenStart = currNode; evenEnd = evenStart; } else { evenEnd -> next = currNode; evenEnd = evenEnd -> next; } } // If current value is odd, add // it to odd values list. else { if(oddStart == NULL) { oddStart = currNode; oddEnd = oddStart; } else { oddEnd -> next = currNode; oddEnd = oddEnd -> next; } } // Move head pointer one step in // forward direction currNode = currNode -> next; } // If either odd list or even list // is empty, no change is required // as all elements are either even // or odd. if(oddStart == NULL || evenStart == NULL) { return; } // Add odd list after even list. evenEnd -> next = oddStart; oddEnd -> next = NULL; // Modify head pointer to // starting of even list. *head_ref = evenStart;}// 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->1->4->6->9->10->11 */ push(&head, 11); push(&head, 10); push(&head, 9); push(&head, 6); push(&head, 4); push(&head, 1); push(&head, 0); printf("Original Linked list "); printList(head); segregateEvenOdd(&head); printf("Modified Linked list "); printList(head); return 0;}// This code is contributed by NIKHIL JINDAL. |
Output:
Original Linked List 0 1 4 6 9 10 11 Modified Linked List 0 4 6 10 1 9 11
Time complexity: O(n)
Please refer complete article on Segregate even and odd nodes in a Linked List for more details!
