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Remove Linked List Elements⚓︎

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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:

  • 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, 10^4].
  • 1 <= Node.val <= 50
  • 0 <= val <= 50

Solution⚓︎

Solution 1⚓︎

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/**
 * 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) {
        // NOTE!!! use "while" here! (not "if")
        while (head != NULL && head->val == val) {
            ListNode *temp = head;
            head = head->next;
            delete temp;
        }

        ListNode *current = head;
        while (current != NULL && current->next != NULL) {
            if (current->next->val == val) {
                ListNode *temp = current->next;
                current->next = current->next->next;
                delete temp;
            } else {
                current = current->next;
            }
        }

        return head;
    }
};
  • Time complexity: \(O(n)\);
  • Space complexity: \(O(1)\)
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/**
 * 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* dummyHead = new ListNode(0);
        dummyHead->next = head;

        ListNode* current = dummyHead;
        while (current->next != NULL) {
            if (current->next->val == val) {
                ListNode *temp = current->next;
                current->next = current->next->next;
                delete temp;
            } else {
                current = current->next;
            }
        }

        // NOTE!!! Do not return head directly (because head may have been deleted)
        head = dummyHead->next;
        delete dummyHead;
        return head;
    }
};

Another way of writing:

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/**
 * 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) {
        auto dummy = new ListNode(-1);
        dummy->next = head;
        for (auto p = dummy; p; p = p->next) {
            auto cur = p->next;
            while (cur && cur->val == val) cur = cur->next;
            p->next = cur;
        }
        return dummy->next;
    }
};

Solution 3: Recursion⚓︎

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/**
 * 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) {
        if (!head) return head;

        head->next = removeElements(head->next, val);
        return head->val == val ? head->next : head;
    }
};