Given the head of a singly 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:
Input: head =[1, 2, 6, 3, 4, 5, 6]
, val = 6 Output:[1, 2, 3, 4, 5]
Your algorithm should run in O(n) time and use O(1) extra space.
The core challenge of this problem is to traverse the linked list and remove nodes that match the given value without using extra space. This is a common problem in linked list manipulation, often encountered in various applications such as memory management, data filtering, and more.
Potential pitfalls include handling the removal of nodes at the beginning of the list and ensuring that the list remains properly linked after node removal.
To solve this problem, we can use a single pass through the linked list, maintaining a reference to the previous node to help with node removal. Here’s a step-by-step approach:
A naive solution would involve creating a new linked list and copying over only the nodes that do not match the given value. However, this approach uses extra space and is not optimal.
We can optimize the solution by modifying the linked list in place. The idea is to use two pointers: one to traverse the list and another to keep track of the last node that does not need to be removed. This way, we can adjust the pointers to skip over nodes that need to be removed.
Here’s a step-by-step breakdown of the optimized algorithm:
class ListNode {
int val;
ListNode next;
ListNode(int x) { val = x; }
}
public class Solution {
public ListNode removeElements(ListNode head, int val) {
// Create a dummy node that points to the head
ListNode dummy = new ListNode(0);
dummy.next = head;
// Initialize current and previous pointers
ListNode current = head;
ListNode prev = dummy;
// Traverse the list
while (current != null) {
if (current.val == val) {
// Skip the current node
prev.next = current.next;
} else {
// Move the previous pointer
prev = current;
}
// Move the current pointer
current = current.next;
}
// Return the new head
return dummy.next;
}
}
The time complexity of this approach is O(n), where n is the number of nodes in the linked list. This is because we traverse the list once. The space complexity is O(1) since we are not using any extra space other than a few pointers.
Potential edge cases include:
Each of these cases is handled by the algorithm as it uses a dummy node and adjusts pointers accordingly.
To test the solution comprehensively, consider the following test cases:
head = []
, val = 1
head = [1, 1, 1]
, val = 1
head = [1, 2, 3]
, val = 4
head = [1, 2, 3]
, val = 1
Use a testing framework like JUnit to automate these tests.
When approaching linked list problems, always consider edge cases and think about how to handle node removal and pointer adjustments. Practice by solving similar problems and studying different algorithms for linked list manipulation.
In this blog post, we discussed how to remove elements from a linked list in O(n) time complexity using Java. We covered the problem definition, approach, algorithm, code implementation, complexity analysis, edge cases, and testing. Understanding and solving such problems is crucial for mastering linked list manipulation and improving problem-solving skills.