Remove Linked List Elements

LeetCode 203 | Difficulty: Easy

Easy

Problem Description

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.

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]`.

- `1 <= Node.val <= 50`

- `0 <= val <= 50`

Topics: Linked List, Recursion

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Approach

Linked List

Use pointer manipulation. Common techniques: dummy head node to simplify edge cases, fast/slow pointers for cycle detection and middle finding, prev/curr/next pattern for reversal.

When to use

In-place list manipulation, cycle detection, merging lists, finding the k-th node.

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Solutions

Solution 1: C# (Best: 172 ms)

Metric	Value
Runtime	172 ms
Memory	N/A
Date	2017-09-22

Solution

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode RemoveElements(ListNode head, int val) {
        if(head==null) return head;
        ListNode dummy = new ListNode(Int32.MinValue);
        dummy.next = head;
        ListNode temp = head;
        ListNode prev = dummy;
        while (temp != null)
        {
            if (temp.val == val)
            {
                prev.next = temp.next;
                temp = temp.next;
                
            }
            else { prev = temp; temp = temp.next; } 
        }
        return dummy.next;
    }
}

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Complexity Analysis

Approach	Time	Space
Linked List	$O(n)$	$O(1)$

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Interview Tips

Key Points

• Start by clarifying edge cases: empty input, single element, all duplicates.
• Draw the pointer changes before coding. A dummy head node simplifies edge cases.