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Delete Node in a Linked List

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 your function.

Example 2:

Input: head = [4,5,1,9], node = 1
Output: [4,5,9]
Explanation: You are given the third node with value 1, the linked list should become 4 -> 5 -> 9 after calling your 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 Explanation for Deleting a Node in a Linked List

This problem presents a unique challenge: deleting a node from a singly linked list without access to the head node. The solution leverages a clever trick to achieve this in constant time and space.

Approach: Data Overwriting

The key insight is that we don't need to physically remove the node from memory; we only need to remove its value from the linked list's sequence. Since we're guaranteed the node isn't the last node, it has a successor. The solution works by copying the value of the next node into the current node, effectively "overwriting" the value we want to delete. Then, we advance the next pointer of the current node to skip over the next node, effectively removing it from the list's sequence.

Algorithm

  1. Copy Value: Copy the value of node.next (the next node) into node.val (the current node).
  2. Skip Node: Update node.next to point to node.next.next (the node after the next node), effectively bypassing the node we want to delete.

Time and Space Complexity

  • Time Complexity: O(1) - The operations (copying a value and updating a pointer) are constant-time regardless of the list's size.
  • Space Complexity: O(1) - The solution uses a constant amount of extra space.

Code Implementation (Python)

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
 
 
class Solution:
    def deleteNode(self, node):
        """
        :type node: ListNode
        :rtype: void Do not return anything, modify node in-place instead.
        """
        node.val = node.next.val  # Copy value from next node
        node.next = node.next.next # Skip the next node

The code in other languages (Java, C++, Go, TypeScript, JavaScript, C#, C) follows the same algorithmic approach, just with syntax specific to each language. They all perform the value copy and pointer update in O(1) time and O(1) space. The crucial part remains consistent across all implementations: directly manipulating the node's data and pointers to achieve the deletion without needing access to the head.