算法

(模拟) $O(n)$

用一个 dummy 节点指向头节点，维护一个 cur 节点，每当他下一个节点为要删除的数时，删除下一个节点，直到当前节点为最后一个节点。

C++ 代码

class Solution {
public:
    ListNode* removeElements(ListNode* head, int val) {
        ListNode dummy;
        dummy.next = head;
        head = &dummy;

        while (head->next != NULL) {
            if (head->next->val == val) {
                head->next = head->next->next;
            } else {
                head = head->next;
            }
        }

        return dummy.next;
    }
};

Java 代码

class Solution {
    public ListNode removeElements(ListNode head, int val) {
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode cur = dummy;

        while (cur.next != null) {
            if (cur.next.val == val) cur.next = cur.next.next;
            else cur = cur.next;
        }

        return dummy.next;
    }
}

Python 代码

class Solution:
    def removeElements(self, head: ListNode, val: int) -> ListNode:
        dummy = ListNode(0)
        dummy.next = head
        cur = dummy
        while cur.next:
            if cur.next.val == val:
                cur.next = cur.next.next
            else:
                cur = cur.next
        return dummy.next