# 203. Remove Linked List Elements

## 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, 104].
• 1 <= Node.val <= 50
• 0 <= val <= 50

## Solutions

• /**
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode() {}
*     ListNode(int val) { this.val = val; }
*     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode removeElements(ListNode head, int val) {
ListNode dummy = new ListNode(-1, head);
ListNode pre = dummy;
while (pre.next != null) {
if (pre.next.val != val)
pre = pre.next;
else
pre.next = pre.next.next;
}
return dummy.next;
}
}

• class Solution {
public:
ListNode* removeElements(ListNode* head, int val) {
ListNode* dummy = new ListNode();
ListNode* p = dummy;
while (p->next) {
if (p->next->val == val) {
p->next = p->next->next;
} else {
p = p->next;
}
}
return dummy->next;
}
};


• # 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:
pre = dummy
while pre.next:
if pre.next.val != val:
pre = pre.next
else:
pre.next = pre.next.next
return dummy.next


• func removeElements(head *ListNode, val int) *ListNode {
dummy := new(ListNode)
p := dummy
for p.Next != nil {
if p.Next.Val == val {
p.Next = p.Next.Next
} else {
p = p.Next
}
}
return dummy.Next
}

• /**
* class ListNode {
*     val: number
*     next: ListNode | null
*     constructor(val?: number, next?: ListNode | null) {
*         this.val = (val===undefined ? 0 : val)
*         this.next = (next===undefined ? null : next)
*     }
* }
*/

function removeElements(head: ListNode | null, val: number): ListNode | null {
const dummy: ListNode = new ListNode(0, head);
let cur: ListNode = dummy;
while (cur.next != null) {
if (cur.next.val === val) {
cur.next = cur.next.next;
} else {
cur = cur.next;
}
}
return dummy.next;
}


• public class Solution {
public ListNode RemoveElements(ListNode head, int val) {
ListNode newTail = null;
while (current != null)
{
if (current.val != val)
{
{
}
else
{
newTail.next = current;
newTail = current;
}
}
current = current.next;
}
if (newTail != null) newTail.next = null;
}
}

• // Definition for singly-linked list.
// #[derive(PartialEq, Eq, Clone, Debug)]
// pub struct ListNode {
//   pub val: i32,
//   pub next: Option<Box<ListNode>>
// }
//
// impl ListNode {
//   #[inline]
//   fn new(val: i32) -> Self {
//     ListNode {
//       next: None,
//       val
//     }
//   }
// }
impl Solution {
pub fn remove_elements(head: Option<Box<ListNode>>, val: i32) -> Option<Box<ListNode>> {
let mut dummy = Box::new(ListNode { val: 0, next: head });
let mut cur = &mut dummy;
while let Some(mut node) = cur.next.take() {
if node.val == val {
cur.next = node.next.take();
} else {
cur.next = Some(node);
cur = cur.next.as_mut().unwrap();
}
}
dummy.next.take()
}
}