Javascript alteration of memory of object when implementing deleting node in linked list

Question

I need some pointer on how does memory in Javascript works. In my implementation of removing the n-th node from the tail of a linked list I have a place holder resultHead.

At the end of the code I have console log the resultHead and the first time is just the original assignment of head:

{"val":1,"next":{"val":2,"next":{"val":3,"next":{"val":4,"next":{"val":5,"next":null}}}}}

But the second time after I have this line:

slow.next = slow.next.next

Then the console logs out:

{"val":1,"next":{"val":2,"next":{"val":3,"next":{"val":5,"next":null}}}}

/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */
/**
 * @param {ListNode} head
 * @param {number} n
 * @return {ListNode}
 */
var removeNthFromEnd = function(head, n) {
    if(head.next === null){return null}
    // use resultHead as a place holder and its next is the actual head
    let resultHead = new ListNode(null);
    resultHead.next = head;
    // find the difference between fast and slow pointer
    let fast = resultHead; 
    let slow = resultHead; 
    for(let i=0; i<=n; i++){
        fast=fast.next;

    }

    while(fast){
        slow = slow.next
        fast = fast.next   
    } 
    console.log(JSON.stringify(resultHead.next))
    slow.next = slow.next.next
    console.log(JSON.stringify(resultHead.next))
    return resultHead.next;
};


removeNthFromEnd({"val":1,"next":{"val":2,"next":{"val":3,"next":{"val":4,"next":{"val":5,"next":null}}}}}, 2)

So the code should remove the node with node.val = 4

How can the slow.next = slow.next.next changed the value of resultHead?

I also tried to console log the resultHead in different places but only after the slow.next = slow.next.next changed the value of resultHead.

chris · Accepted Answer · 2019-03-24 04:23:46Z

2

Remember, javascript passes objects by reference. resultHead.next = head means modifying resultHead.next modifies head externally. You set slow and fast to resultHead so modifying slow.next modifies head.

i.e.

slow === resultHead
resultHead.next === head
slow.next === head

edited Mar 24, 2019 at 4:23

answered Mar 24, 2019 at 4:14

chris

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1 Comment

Ming Huang Over a year ago

Okay so what happens in the while loop when I set slow=slow.next. Doesn't it set the slow variable to point to a different object(the next node in the list)? After the loop ends it will end at node.val=3, now does the slow takes the node reference inside the resultHead?

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