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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

Solution 1

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# 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
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/**
 * Definition for singly-linked list.
 * 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;
    }
}
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class Solution {
public:
    ListNode* removeElements(ListNode* head, int val) {
        ListNode* dummy = new ListNode();
        dummy->next = head;
        ListNode* p = dummy;
        while (p->next) {
            if (p->next->val == val) {
                p->next = p->next->next;
            } else {
                p = p->next;
            }
        }
        return dummy->next;
    }
};
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func removeElements(head *ListNode, val int) *ListNode {
    dummy := new(ListNode)
    dummy.Next = head
    p := dummy
    for p.Next != nil {
        if p.Next.Val == val {
            p.Next = p.Next.Next
        } else {
            p = p.Next
        }
    }
    return dummy.Next
}
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/**
 * Definition for singly-linked list.
 * 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;
}
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// 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 {