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javapointersmethodslinked-list
brain storm
brain storm

Reputation: 31252

Remove method for linkedList implementation in Java

I have this method from lecture on removing elements from linkedList at specified index. I understand how the method works, but I do not understand why the for-loop leaves the current node pointer two index before the desired index.

Here is the method:

public void remove(int index) {
        if (index == 0) {
            // removing the first element must be handled specially
            front = front.next;
        } else {
            // removing some element further down in the list;
            // traverse to the node before the one we want to remove
            ListNode current = front;
            for (int i = 0; i < index - 1; i++) {
                current = current.next;
            }

            // change its next pointer to skip past the offending node
            current.next = current.next.next;
        }
    }

The for-loop goes from 0 to < index-1, while I thought it should go from 0 to < index. In this way, the pointer is at one index before the index that needed to be deleted. However, the above method works fine.

For eg: in the below LinkedList enter image description here

Lets consider removing Node C. By the above loop-construct, current pointer will be pointing at Node A and current.next will be Node B. current.next.next will be Node C. Doing current.next=current.next.next will result in Node B deletion rather than Node C.

I think something is wrong with my understanding, can somebody explain?

Upvotes: 2

Views: 22409

Answers (1)

Sotirios Delimanolis
Sotirios Delimanolis

Reputation: 279990

The for-loop goes from 0 to < index-1

In your example, removing C means index is 2. So i only goes to 0, since 1 is not < 1.

current starts at A, the for loops once and current goes to B.

current is B, so current.next.next is D, which effectively removes C.

Upvotes: 2

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