JuneChallenge (63) | Only access a Node, detete It

Delete Node in a Linked List

Write a function to delete a node (except the tail) in a singly linked list, given only access to that node.

Given linked list -- head = [4,5,1,9], which looks like following:

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.

Note:

• The linked list will have at least two elements.
• All of the nodes' values will be unique.
• The given node will not be the tail and it will always be a valid node of the linked list.
• Do not return anything from your function.

In a linked list, you can only accest one node, so, you can't know the node prev before.

We alway delete a node in linkedlist like: prev -> node -> node.next. We want to delete node:

prev.next = node.next;

This problem is possible because you don't know the prev.next. So, we leak a little, instead of delete a node like past, we will delete the next node.

Example: 4 -> 5 -> 1 -> 9. We need delete node (5): node(5).val = node(5).next.val = 1: 4->1->1->9. node(5).next = node(5).next.next: 5->1->9. Done !
Thank you, and see you later.