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

Code

Indirect Pointer

Time Complexity: $O(N)$, Space Complexity: $O(1)$

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* removeElements(struct ListNode* head, int val) {
    if(!head) return NULL;
    struct ListNode** indirect = &head;
    while(*indirect) {
        if((*indirect)->val == val) {
            struct ListNode* to_be_free = *indirect;
            *indirect = (*indirect)->next;
            free(to_be_free);
        } else {
            indirect = &((*indirect)->next);
        }
    }    
    return head;
}

Pointer

Time Complexity: $O(N)$, Space Complexity: $O(1)$

/**
 * 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* dummy = new ListNode();
        dummy->next = head;
        ListNode* curr = dummy;
        while(curr->next) {
            if(curr->next->val == val) {
                curr->next = curr->next->next;
            } else {
                curr = curr->next;
            }    
        }
        return dummy->next;
    }
};

Source

• Remove Linked List Elements - LeetCode