Reverse a linkedlist using javascript

Question

This is a singly linked list and i want to reverse it

I found this in stackoverflow but it doesn't help me

it returns 1 not [16,5,10,1]

What i know about reverse a linkedlist is to

let the first node point to the null

the second node point to the first node

the third node point to the second node

Could someone help me to figure out how to reverse a linkedlist in my code?

Here is my JS:

class LinkedList {
  constructor(value) {
    this.head = {
      value: value,
      next: null
    };
    this.tail = this.head;
    this.length = 1;
  }
  append(value) {
    const newNode = {
      value: value,
      next: null
    }
    this.tail.next = newNode;
    this.tail = newNode;
    this.length++;
    return this;
  }
  prepend(value) {
    const newNode = {
      value: value,
      next: null
    }
    newNode.next = this.head;
    this.head = newNode;
    this.length++;
    return this;
  }
  printList() {
    const array = [];
    let currentNode = this.head;
    while (currentNode !== null) {
      array.push(currentNode.value)
      currentNode = currentNode.next
    }
    return array;
  }
  insert(index, value) {
    //Check for proper parameters;
    if (index >= this.length) {
      console.log('yes')
      return this.append(value);
    }

    const newNode = {
      value: value,
      next: null
    }
    const leader = this.traverseToIndex(index - 1);
    const holdingPointer = leader.next;
    leader.next = newNode;
    newNode.next = holdingPointer;
    this.length++;
    return this.printList();
  }
  traverseToIndex(index) {
    //Check parameters
    let counter = 0;
    let currentNode = this.head;
    while (counter !== index) {
      currentNode = currentNode.next;
      counter++;
    }
    return currentNode;
  }
  remove(index) {
    // Check Parameters      
    const leader = this.traverseToIndex(index - 1);
    const unwantedNode = leader.next;
    leader.next = unwantedNode.next;
    this.length--;
    return this.printList();
  }
  reverse() {
    let currentNode = this.head;
    var previous = null;

    while (currentNode) {
      // save next or you lose it!!!
      var save = currentNode.next;
      // reverse pointer
      currentNode.next = previous;
      // increment previous to current node
      previous = currentNode;
      // increment node to next node or null at end of list
      currentNode = save;
    }
    return this.printList()
  }
}

let myLinkedList = new LinkedList(10);
myLinkedList.append(5)
myLinkedList.append(16)
myLinkedList.prepend(1)
myLinkedList.insert(2, 99)
myLinkedList.remove(2)
myLinkedList.reverse() //should return [16,5,10,1]

Jonas Wilms · Accepted Answer · 2019-03-06 06:27:11Z

2

You are actually quite close, after your loop just re set the head & tail:

 while(/*...*/ { /*...*/ }

 this.tail = this.head;
 this.head = previous;

answered Mar 6, 2019 at 6:27

Jonas Wilms

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Ajay yadav · Accepted Answer · 2019-03-06 06:18:25Z

0

function reversesll(sll){

  if(!sll.head || !sll.head.next) return sll;

  var nodes=[], 
    current = sll.head;
  //storing all the nodes in an array
  while(current){
    nodes.push(current);
    current = current.next;
  }

  var reversedLL = new LinkedList();

  reversedLL.head = nodes.pop();
  current = reversedLL.head;

  var node = nodes.pop();  
  //make sure to make next of the newly inserted node to be null
  //other wise the last node of your SLL will retain its old next.
  while(node){
     node.next = null;
     current.next = node;

     current = current.next;
     node = nodes.pop();
  }
  return reversedLL;
}

//create the LL
var sll = new LinkedList();
sll.push(1);
sll.push(2);
sll.push(3);
sll.push(4);
sll.push(5);

//test it
reversesll(sll);
//{head: {value:5, next:{value: 4, next: {value: 3, next: {value:2, next:{value:1, next: null}}}}}}

answered Mar 6, 2019 at 6:18

Ajay yadav

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2 Comments

Jacky Over a year ago

looks like you miss my point. I have already had a reverse function in my class.Just don't create a new one

Ajay yadav Over a year ago

let me check again

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