1

I am trying to solve Flatten a Multilevel Doubly Linked List problem on a leetcode. I am able to create a singly linked list from an array but don't know how to create a doubly linked list.

You are given a doubly linked list, which contains nodes that have a next pointer, a previous pointer, and an additional child pointer. This child pointer may or may not point to a separate doubly linked list, also containing these special nodes. These child lists may have one or more children of their own, and so on, to produce a multilevel data structure as shown in the example below.

Given the head of the first level of the list, flatten the list so that all the nodes appear in a single-level, doubly linked list. Let curr be a node with a child list. The nodes in the child list should appear after curr and before curr.next in the flattened list.

Return the head of the flattened list. The nodes in the list must have all of their child pointers set to null.

Ques link: Flatten a Multilevel Doubly Linked List

My code:

/**
 * // Definition for a Node.
 * function Node(val,prev,next,child) {
 *    this.val = val;
 *    this.prev = prev;
 *    this.next = next;
 *    this.child = child;
 * };
 */

/**
 * @param {Node} head
 * @return {Node}
 */
var flatten = function(head) {
    let array=[];
     let list=head;   
const childAdd=(node)=>{
       while(node){  
          array.push(node.val);
        if(node.child)
            childAdd(node.child);

         node=node.next;
    }
    } 
       while(list){
        array.push(list.val);
        if(list.child)
            childAdd(list.child);
           
           list=list.next;
       }
     console.log(array)
       let resultList=null
      
       for(let i=array.length-1; i>=0; i--){
           resultList={
               val:array[i],
               next:resultList,
           }
 let prev=null;
    while(head){
        delete head.child
        resultList.prev=prev;
        prev=head;
        head=head.next;
    }
    console.log(resultList)
       return resultList;
};

Output:

[
   1,  2, 3,  7, 8,
  11, 12, 9, 10, 4,
   5,  6
]
{
  val: 1,
  next: { val: 2, next: { val: 3, next: [Object] } },
  prev: <ref *1> {
    val: 5,
    prev: { val: 4, prev: [Object], next: [Circular *1] },
    next: { val: 6, prev: [Circular *1], next: null }
  }
}

Expected Output:

The linked list [1,2,3,7,8,11,12,9,10,4,5,6] is not a valid doubly linked list.

Expected: [1,2,3,7,8,11,12,9,10,4,5,6]

How can I add prev which points to the previous node of the list in my list resultList?

PS: I don't want a solution to this problem instead I want to know how can I create a doubly linked list from an array.

Improve this question
2
  • Although you can solve this by copying data into an array, the real challenge is to do this without extra storage -- without creating arrays. Commented Aug 19, 2022 at 18:42
  • Your code has an unclosed brace... looks like the for loop is missing one. Commented Aug 19, 2022 at 18:47

4 Answers 4

Reset to default
1

LeetCode expects you to return a Node instance, i.e. an object with four properties (also including child).

And the prev property can be set correctly when you set the next: that next node's prev property should become a back reference to the object of which you set the next property.

So replace this block of code (which misses a closing brace):

    for(let i=array.length-1; i>=0; i--){
        resultList={
            val:array[i],
            next:resultList,
        }
    let prev=null;
    while(head){
        delete head.child
        resultList.prev=prev;
        prev=head;
        head=head.next;
    }

With this:

    for (let i = array.length - 1; i >= 0; i--) {
        resultList = new Node(array[i], null, resultList, null);
        if (resultList.next) resultList.next.prev = resultList;
    }

Other remark

This solution will pass the tests, but it is not very efficient. It uses O(n) extra space, while this can be done with O(1) extra space:

  • Avoid creating arrays
  • Avoid creating new Node instances -- reuse the ones you already have
Improve this answer
Sign up to request clarification or add additional context in comments.

2 Comments

resultList object I have created in the loop is same which new Node object is. So why you have used new Node? Also just removing while loop and adding resultList.next.prev = resultList in my code solves the problem.
Yes, it also works when you create new objects with the object literal that you used (and when you add the child property in it), but since LeetCode uses this constructor, it is quite natural to use the same constructor (Node) instead of creating these objects with the default constructor (Object).
1 

    function arrayToDoublyLinkedList(array) {
        let curr;
        let next = null;
        let prev = null;
    
        // go backwards so we end up with curr being the head
        for(let i = array.length - 1; i >= 0; --i) {
            curr = {
                val: array[i],
                prev: null, // except for the head, will set to a node on the next iteration
                next: next,
                child: null, // required to be null by the leetcode problem statement
            };
            
            if (null !== next) {
                next.prev = curr;
            }
            
            next = curr;
        }
        
        return curr;
    }
    
    console.log(arrayToDoublyLinkedList([1,2,3,4,5,6,7,8]));

Improve this answer

Comments

0

The hardest part is creating the list from the arr. Basically we iterate the array creating nodes, connect node to previous and connect previous to node. The trick here is when encountering null is to set a pointer (bol) to beginning of current list then while still encountering null we increment the bol. Finally when we do find a value we know to create a node that is a child of bol.

For completeness I added the solution (which is the easy part) but feel free to find it yourself.

function Node(val, prev, next, child) {
  this.val = val;
  this.prev = prev;
  this.next = next;
  this.child = child;
};


function create_list(arr) {
  var prev = null;
  var first = null;
  var bol = null;
  var result = null;

  while (arr.length) {
    var val = arr.shift();
    if (val === null) {
      if (bol) {
        bol = bol.next;
      } else {
        bol = first;
        prev = null;
      }
    } else {
      var node = new Node(val, prev, null, null);
      if (!result) {
        result = node;
      }
      if (bol) {
        bol.child = node;
        bol = null;
      }
      if (node.prev) {
        node.prev.next = node;
      } else {
        first = node;
      }
      prev = node;
    }
  }
  return result;
}

var head = [1, 2, 3, 4, 5, 6, null, null, null, 7, 8, 9, 10, null, null, 11, 12];
var copy = [...head];
var list = create_list(head);

function flatten(list) {
  var arr = [];
  while (list) {
    arr.push(list.val);
    if (list.child) {
      var brr = flatten(list.child);
      arr = arr.concat(brr);
    }
    list = list.next;
  }
  return arr;
}
console.log("list as array: " + JSON.stringify(copy))
console.log("flattened list: " + JSON.stringify(flatten(list)))

console.log("example 2: " + JSON.stringify(flatten(create_list([1,2,null,3]))))
console.log("example 3: " + JSON.stringify(flatten(create_list([]))))
.as-console-wrapper {
  max-height: 100% !important;
}

Improve this answer

Comments

0

A rather short version:

function llist(arr){
  function reduce(acc, val){
    return val ? (val.prev = acc, reduce(val, val.next)) : acc;
  }
  return reduce(null, arr.reduceRight((next, value) => ({value, next}), null));
}
var arr = [1, 2, 3, 4 ,5];
console.log(llist(arr));
Improve this answer

1 Comment

i ended up with a reversed, singly linked list. node.next never worked but node.prev did

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.