Given the head of a sorted linked list, delete all duplicates such that each element appears only once. Return the linked list sorted as well.

Input: head = [1,1,2] Output: [1,2]

Solution:

    let curr=head;
    while(curr && curr.next){
        
        if(curr.val===curr.next.val){
            curr.next=curr.next.next;
        }
        else{
            curr=curr.next;
        }
    }
 return head;

I am trying to solve this question on leetcode and found one solution which I am not able to understand one thing, Why are we taking the head in let curr? And, If I am trying to do the same thing without takinghead in another variable then I am getting only [2] as output.

CodePudding user response：

Why are we taking the head in let curr?

For two reasons:

1. curr needs to be initialised to something, otherwise it will have an undefined value, making the while condition always false, so there will be no iterations and no removals happening.

2. curr is intended to refer to each (non-duplicate) node one after the other, so it makes sense to start with the first node, which is head.

And, If I am trying to do the same thing without taking head in another variable then I am getting only [2] as output.

That could only happen when you also change the return head by return curr. If you do that, then you will always return a list that has no more than one node, because after the loop has completed, curr will refer to the last node in the list (if it is not empty).

In order to return all nodes in a list, you need to always return its first node, and that is what head represents.