Problem : Easy

LeetCode: Given 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.

Image from Leetcode

Animation from Zybooks

Thought

• I have tried to implement this in C++ before but this is my first time using Java to solve it.
• Rather than using pointer, Java ListNode has a filed Next as a reference point to the next value in the linked list.
• Useful youtube link #5 Linked List Implementation in Java Part 1 Data Structures
• Another useful link in Chinese from Carl 手把手带你学会操作链表
• I have tried to understand this data staructure by drawing on the paper with understanding of memory location usesage.
• Using dummy node can keep consistance of treversing to remove target elments from the list. In this case, head will stay the same, instaed dummy and currNode will keep traversing until the end of the list.
• Hint for this challenge will be returning dummy.next;

Steps:

• Check if the head node exsits, if not return head;
• Initialise a new node: dummy node
• Dummy will point to the head
• Initialise another node as currNode = dummy
• Using while loop (Condition is currNode.next != null) to loop until find the target value
• When loop finishes(when currNode point to null), return dummy.next

Solution: Java

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */

class Solution {
    public ListNode removeElements(ListNode head, int val) {

        if(head == null){
            return head;
        }

        ListNode dummy = new ListNode();
        dummy.next = head;
        ListNode currNode = dummy;

        while (currNode.next != null ){
            if(currNode.next.val == val) {
                currNode.next = currNode.next.next;

            }
            else{
                currNode = currNode.next;
            }
        }
        return dummy.next;    
    }
}

For more instructions head over to the LeetCode.