LeetCode Daily: Delete Nodes from Linked List Present in Array Solution in Java | September 6, 2024

🔍 LeetCode Problem of the Day: Delete Nodes from Linked List Present in Array

In today's LeetCode daily challenge, we solve the problem of removing nodes from a linked list if their values exist in a given array, using a HashSet for efficient lookups.

👉 Solution: Pinned on the comments

🌟 Problem Description:
Given the head of a linked list and an array of integers nums, the task is to remove all nodes from the linked list that contain values present in the nums array. The list should be modified in-place.

🔑 Key Points:
HashSet for Fast Lookup: We use a HashSet to store the elements of the array for O(1) lookups.
Dummy Node Technique: A dummy node is used to simplify edge cases where deletions happen at the head of the linked list.
Two Pointer Approach: We use a prev and current pointer to traverse the list and delete nodes by adjusting the next pointers.

📝 Code Explanation:
Set Construction: We first store all the numbers from nums in a HashSet for quick access.
Dummy Node: A dummy node is introduced to handle edge cases, such as deletions at the head.

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[AlgoXploration]

class Solution {
public ListNode modifiedList(int[] nums, ListNode head) {
HashSet<Integer> set = new HashSet<>();
for (int num : nums)
set.add(num);

ListNode dummy = new ListNode(0);
dummy.next = head;
ListNode current = head;
ListNode prev = dummy;

while (current != null) {
if (set.contains(current.val))
prev.next = current.next;
else
prev = current;
current = current.next;
}

return dummy.next;
}
}