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237. 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.

[TOC]

Solution

Overview

The problem is an extension of LeetCode 203. Remove Linked List Elements.

In the original LeetCode 203. Remove Linked List Elements, we were given the head of the linked list, so we can go to the desired node and connect the previous nodes to the next node.

However, here we are not given access to the head of the linked list, and it makes the problem a bit tricky.

Approach 1: Delete next node instead of current one

Intuition

To solve the problem, let's look at the condition carefully,

It is guaranteed that the node to be deleted is not a tail node in the list.

There are a few observations here,

The conventional deletion approach will fail here since we are not able to get the previous node by iterating from the head of the linked list. In fact, we do not have any method to fetch the previous node.

Without the knowledge of the previous node, it's not possible to delete the current node.

Notice that we are told the given node is not a tail node. Therefore, we can delete the next node instead of the current node given, and "pretend" that the node we are given is the next node.

Using this intuition, let's understand how to implement this problem.

Algorithm

By analyzing the above two key observations, we can derive the following algorithm,

Store the next node in a temporary variable.

Copy data of the next node to the current node.

Change the next pointer of the current node to the next pointer of the next node.

Note: Above 3 steps makes sure that your current node becomes same as next node and then you can safely delete next node from the Linked List

Steps

Initial Linked List

Update current node with next node details

Update current node next pointer with next node next pointer

Implementation

Complexity Analysis

Time Complexity: $O(1)$ since only 1 node needs to be updated and we only traverse one node behind.

Space Complexity: $O(1)$, since we use constant extra space to track the next node.

https://leetcode.com/articles/delete-nod...nked-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.

[TOC]

Solution

Overview

The problem is an extension of LeetCode 203. Remove Linked List Elements.

In the original LeetCode 203. Remove Linked List Elements, we were given the head of the linked list, so we can go to the desired node and connect the previous nodes to the next node.

However, here we are not given access to the head of the linked list, and it makes the problem a bit tricky.

Approach 1: Delete next node instead of current one

Intuition

To solve the problem, let's look at the condition carefully,

It is guaranteed that the node to be deleted is not a tail node in the list.

There are a few observations here,

The conventional deletion approach will fail here since we are not able to get the previous node by iterating from the head of the linked list. In fact, we do not have any method to fetch the previous node.

Without the knowledge of the previous node, it's not possible to delete the current node.

Notice that we are told the given node is not a tail node. Therefore, we can delete the next node instead of the current node given, and "pretend" that the node we are given is the next node.

Using this intuition, let's understand how to implement this problem.

Algorithm

By analyzing the above two key observations, we can derive the following algorithm,

Store the next node in a temporary variable.

Copy data of the next node to the current node.

Change the next pointer of the current node to the next pointer of the next node.

Note: Above 3 steps makes sure that your current node becomes same as next node and then you can safely delete next node from the Linked List

Steps

Initial Linked List

Update current node with next node details

Update current node next pointer with next node next pointer

Implementation

Complexity Analysis

Time Complexity: $O(1)$ since only 1 node needs to be updated and we only traverse one node behind.

Space Complexity: $O(1)$, since we use constant extra space to track the next node.

https://leetcode.com/articles/delete-nod...nked-list/