# LeetCode 237. Delete Node in a Linked List

## Description

Write a function to delete a node in a singly-linked list. You will not be given access to the `head` of the list, instead you will be given access to the node to be deleted directly.

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

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.
```

Example 3:

```Input: head = [1,2,3,4], node = 3
Output: [1,2,4]
```

Example 4:

```Input: head = [0,1], node = 0
Output: 
```

Example 5:

```Input: head = [-3,5,-99], node = -3
Output: [5,-99]
```

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

## Explanation

Simply make the deleting node to be the next node.

## Python Solution

``````# 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
node.next = node.next.next

``````
• Time complexity: O(1).
• Space complexity: O(1).

## 2 Thoughts to “LeetCode 237. Delete Node in a Linked List”

1. Vickey says:

/**
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public void deleteNode(ListNode node) {

node.val = node.next.val;
node.next = node.next.next;

}
}