Remove Linked List Elements

 iven the head of a linked list and an integer val, remove all the nodes of the linked list that has Node.val == val, and return the new head.

 

Example 1:

Input: head = [1,2,6,3,4,5,6], val = 6

Output: [1,2,3,4,5]

Example 2:

Input: head = [], val = 1

Output: []

Example 3:

Input: head = [7,7,7,7], val = 7

Output: []

 

Constraints:

The number of nodes in the list is in the range [0, 10^4].

We'll use a dummy node to handle edge cases like removing the head.

Steps:

1. Create a dummy node (dummy) before the head.

2. Use a current pointer starting from dummy.

3. Loop while current.next is not null:

   • If current.next.val == val, skip it using current.next = current.next.next

   • Else, move forward.

4. Return dummy.next (new head).

class ListNode {

    int val;

    ListNode next;

    ListNode(int val) {

        this.val = val;

    }

}

public class Solution {

    public ListNode removeElements(ListNode head, int val) {

        // Step 1: Create a dummy node

        ListNode dummy = new ListNode(-1);

        dummy.next = head;

        // Step 2: Initialize current pointer

        ListNode current = dummy;

        // Step 3: Traverse and remove matching nodes

        while (current.next != null) {

            if (current.next.val == val) {

                current.next = current.next.next; // skip node

            } else {

                current = current.next; // move ahead

            }

        }

        // Step 4: Return new head

        return dummy.next;

    }

}