203_Removed-Linked-List-Elements
Description
Remove all elements from a linked list of integers that have value val.
Example:
Input: 1->2->6->3->4->5->6, val = 6
Output: 1->2->3->4->5
Solution
Java solution 1
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode removeElements(ListNode head, int val) {
while (head != null && head.val == val) {
head = head.next;
}
if (head == null) {
return null;
}
ListNode prev = head;
while (prev.next != null) {
if (prev.next.val == val) {
prev.next = prev.next.next;
} else {
prev = prev.next;
}
}
return head;
}
}
Runtime: 7 ms.
Java solution 2
Using dummy head node.
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode removeElements(ListNode head, int val) {
ListNode dummyHead = new ListNode(-1);
dummyHead.next = head;
ListNode prev = dummyHead;
while (prev.next != null) {
if (prev.next.val == val) {
prev.next = prev.next.next;
} else {
prev = prev.next;
}
}
return dummyHead.next;
}
}
Runtime: 8 ms.
Python solution
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def removeElements(self, head, val):
"""
:type head: ListNode
:type val: int
:rtype: ListNode
"""
dummy_head = ListNode(-1)
dummy_head.next = head
prev = dummy_head
while prev.next is not None:
if prev.next.val == val:
prev.next = prev.next.next
else:
prev = prev.next
return dummy_head.next
Runtime: 88 ms. Your runtime beats 74.70 % of python3 submissions.