Leetcode #203: Remove Linked List Elements
In this guide, we solve Leetcode #203 Remove Linked List Elements in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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, 104].
Quick Facts
- Difficulty: Easy
- Premium: No
- Tags: Recursion, Linked List
Intuition
Linked list problems often require pointer manipulation rather than extra memory.
Two-pointer techniques expose cycles, midpoints, or reordering patterns.
Approach
Traverse with fast/slow pointers or reverse sublists when needed.
Maintain invariants carefully to avoid losing nodes.
Steps:
- Traverse with pointers.
- Reverse or split if required.
- Reconnect nodes correctly.
Example
Input: head = [1,2,6,3,4,5,6], val = 6
Output: [1,2,3,4,5]
Python Solution
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def removeElements(self, head: ListNode, val: int) -> ListNode:
dummy = ListNode(-1, head)
pre = dummy
while pre.next:
if pre.next.val != val:
pre = pre.next
else:
pre.next = pre.next.next
return dummy.next
Complexity
The time complexity is O(n). The space complexity is O(1).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.