"reverse a linked list" Code Answer's

Reverse a linked list c++

By Itchy Impala on Jun 20, 2021

#include<bits/stdc++.h>
 
using namespace std;
 
struct node {
    int data;
    struct node *next;
};
 
// To create a demo we have to construct a linked list and this 
// function is to push the elements to the list. 
void push(struct node **head_ref, int data) {
    struct node *node;
    node = (struct node*)malloc(sizeof(struct node));
    node->data = data;
    node->next = (*head_ref);
    (*head_ref) = node;
}
 
// Function to reverse the list
void reverse(struct node **head_ref) {
    struct node *temp = NULL;
    struct node *prev = NULL;
    struct node *current = (*head_ref);
    while(current != NULL) {
        temp = current->next;
        current->next = prev;
        prev = current;
        current = temp;
    }
    (*head_ref) = prev;
}
 
// To check our program 
void printnodes(struct node *head) {
    while(head != NULL) {
        cout<<head->data<<" ";
        head = head->next;
    }
}
 
// Driver function
int main() {
    struct node *head = NULL;
    push(&head, 0);
    push(&head, 1);
    push(&head, 8);
    push(&head, 0);
    push(&head, 4);
    push(&head, 10);
    cout << "Linked List Before Reversing" << endl;
    printnodes(head);
    reverse(&head);
    cout << endl;
    cout << "Linked List After Reversing"<<endl;
    printnodes(head);
    return 0;
}

Source: favtutor.com

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reverse linked list in java to get both head and tail

By Ak Shaw on May 26, 2020

/*
public class ListNode {
    public int val;
    public ListNode next;
    public ListNode(int x) { val = x; next = null; }
}
*/

public static ListNode[] reverse_linked_list(ListNode head) {

        ListNode prev = null;
        ListNode current = head;
        ListNode next;

        ListNode tail = head;

        while (current != null) {

            next = current.next;
            current.next = prev;
            prev = current;
            current = next;
        }

        head = prev;

        ListNode[] result = {head, tail};

        return result;
}

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reverse a linked list

By Saan on Sep 08, 2020

class recursion { 
	static Node head; // head of list 
	static class Node { 
		int data; 
		Node next; 
		Node(int d) 
		{   data = d; 
			next = null; 	} } 
	static Node reverse(Node head) 
	{ 
		if (head == null || head.next == null) 
			return head; 
		/* reverse the rest list and put the first element 
        at the end */
		Node rest = reverse(head.next); 
		head.next.next = head; 
		/* tricky step -- see the diagram */
    	head.next = null; 
		/* fix the head pointer */
		return rest; 
	}  /* Function to print linked list */
	static void print() 
	{ 
		Node temp = head; 
		while (temp != null) { 
			System.out.print(temp.data + " "); 
			temp = temp.next; 
		} 
		System.out.println(); 
	} 
	static void push(int data) 
	{ 
		Node temp = new Node(data); 
		temp.next = head; 
		head = temp; 
	} /* Driver program to test above function*/
public static void main(String args[]) 
{ 
	/* Start with the empty list */
	push(20); 
	push(4); 
	push(15); 
	push(85); 
	System.out.println("Given linked list"); 
	print(); 
	head = reverse(head); 
	System.out.println("Reversed Linked list"); 
	print(); 
} } // This code is contributed by Prakhar Agarwal 

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Revese a Linked List

By Wandering Wolverine on May 28, 2021

LinkedList<Integer> ll = new LinkedList<>();

ll.add(1);
ll.add(2);
ll.add(3);
System.out.println(ll);

LinkedList<Integer> ll1 = new LinkedList<>();

ll.descendingIterator().forEachRemaining(ll1::add);

System.out.println(ll1);

Source: www.journaldev.com

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reverse linkedlist

By Wandering Wolverine on May 13, 2021

Collections.reverse(list);

Source: stackoverflow.com

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linked list reverse

By Udituk on Jun 17, 2021

// using iterative method to reverse linked list in JavaScript
// time complexity: O(n) & space complexity: O(1)
reverse() {
      if (!this.head.next) {
        return this.head;
      }
      
      let prevNode = null;
      let currNode = this.head;
      let nextNode = this.head;
      while(nextNode){
        nextNode = currNode.next;
        currNode.next = prevNode;
        prevNode = currNode;
        currNode = nextNode;
      }
      this.head = prevNode;
      return this.printList();
    }

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