203. Remove Linked List Elements LeetCode Solution
In this guide, you will get 203. Remove Linked List Elements LeetCode Solution with the best time and space complexity. The solution to Remove Linked List Elements problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
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].
1 <= Node.val <= 50
0 <= val <= 50
Complexity Analysis
Time Complexity: O(n)
Space Complexity: O(1)
203. Remove Linked List Elements LeetCode Solution in C++
class Solution {
public:
ListNode* removeElements(ListNode* head, int val) {
ListNode dummy(0, head);
ListNode* prev = &dummy;
for (; head; head = head->next)
if (head->val != val) {
prev->next = head;
prev = prev->next;
}
prev->next = nullptr; // In case that the last value equals `val`.
return dummy.next;
}
}; /* code provided by PROGIEZ */
203. Remove Linked List Elements LeetCode Solution in Java
class Solution {
public ListNode removeElements(ListNode head, int val) {
ListNode dummy = new ListNode(0, head);
ListNode prev = dummy;
for (; head != null; head = head.next)
if (head.val != val) {
prev.next = head;
prev = prev.next;
}
prev.next = null; // In case that the last value equals `val`.
return dummy.next;
}
} // code provided by PROGIEZ
203. Remove Linked List Elements LeetCode Solution in Python
class Solution:
def removeElements(self, head: ListNode, val: int) -> ListNode:
dummy = ListNode(0, head)
prev = dummy
while head:
if head.val != val:
prev.next = head
prev = prev.next
head = head.next
prev.next = None
return dummy.next # code by PROGIEZ
