2024-05-06: LC 237. Delete Node in a Linked List

Difficulty: Middle

 

There is a singly-linked list head and we want to delete a node node in it.

You are given the node to be deleted node. You will not be given access to the first node of head.

All the values of the linked list are unique, and it is guaranteed that the given node node is not the last node in the linked list.

Delete the given node. Note that by deleting the node, we do not mean removing it from memory. We mean:

• The value of the given node should not exist in the linked list.
• The number of nodes in the linked list should decrease by one.
• All the values before node should be in the same order.
• All the values after node should be in the same order.

Custom testing:

• For the input, you should provide the entire linked list head and the node to be given node. node should not be the last node of the list and should be an actual node in the list.
• We will build the linked list and pass the node to your function.
• The output will be the entire list after calling your function.

Example 1:

• Input: head = [4, 5, 1, 9], node = 5
• Output: [4, 1, 9]
• Explanation: You are given the second node with value 5, the linked list should become 4 → 1 → 9 after calling the function

Example 2:

• Input: head = [4, 5, 1, 9], node = 1
• Output: [4, 5, 9]
• Explanation: It is given that the third node with value 1, the linked list should become 4 → 5 → 9 after calling the function

Constraints:

• The number of the nodes in the given list is in the range [2, 1000].
• 1000 <= Node.val <= 1000
• The value of each node in the list is unique.
• The node to be deleted is in the list and is not a tail node.

Solution

• It might seem weird that we do not have access to the head node in the linked list. However, we can view the question into the different angle.
• Anyway the value stored in the node is important part in the linked list and it is pretty easy to copy the information from the next node to the current node.
  • i.e. only one variable is stored in each node, i.e. node.val.
• Final Code

    # Definition for singly-linked list.
    # class ListNode:
    #     def __init__(self, x):
    #         self.val = x
    #         self.next = None

    class Solution:
        def deleteNode(self, node):
            node.val = node.next.val
            node.next = node.next.next
            """
            :type node: ListNode
            :rtype: void Do not return anything, modify node in-place instead.
            """

Analysis

• Time Complexity: O(1), we only store the value of the next node into the current node and set curr.next to curr.next.next.
• Space Complextity: O(1)