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203. Remove Linked List Elements

Difficulty Topics

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

Solution

remove-linked-list-elements.py
# 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: Optional[ListNode], val: int) -> Optional[ListNode]:
        res = curr = ListNode(-1, head)

        while curr:
            while curr.next and curr.next.val == val:
                curr.next = curr.next.next

            curr = curr.next

        return res.next
remove-linked-list-elements.cpp
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* removeElements(ListNode* head, int val) {
        ListNode *curr = new ListNode;
        curr->next = head;
        ListNode *res = curr;

        while (curr){
            while (curr->next && curr->next->val == val)
                curr->next = curr->next->next;
            curr = curr->next;
        }

        return res->next;
    }
};