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Update solution 0237
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# [237. Delete Node in a Linked List](https://leetcode.com/problems/delete-node-in-a-linked-list/)
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## 题目
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Write a function to delete a node (except the tail) in a singly linked list, given only access to that node.
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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.
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Given linked list -- head = [4,5,1,9], which looks like following:
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It is **guaranteed** that the node to be deleted is **not a tail node** in the list.
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**Example 1:**
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Example 1:
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```c
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```
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Input: head = [4,5,1,9], node = 5
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Output: [4,1,9]
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Explanation: You are given the second node with value 5, the linked list should become 4 -> 1 -> 9 after calling your function.
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Explanation:You are given the second node with value 5, the linked list should become 4 -> 1 -> 9 after calling your function.
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```
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Example 2:
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**Example 2:**
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```c
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```
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Input: head = [4,5,1,9], node = 1
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Output: [4,5,9]
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Explanation: You are given the third node with value 1, the linked list should become 4 -> 5 -> 9 after calling your function.
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Explanation:You are given the third node with value 1, the linked list should become 4 -> 5 -> 9 after calling your function.
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```
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Note:
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**Example 3:**
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- The linked list will have at least two elements.
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- All of the nodes' values will be unique.
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- The given node will not be the tail and it will always be a valid node of the linked list.
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- Do not return anything from your function.
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```
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Input: head = [1,2,3,4], node = 3
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Output: [1,2,4]
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```
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**Example 4:**
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```
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Input: head = [0,1], node = 0
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Output: [1]
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```
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**Example 5:**
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```
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Input: head = [-3,5,-99], node = -3
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Output: [5,-99]
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```
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**Constraints:**
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- The number of the nodes in the given list is in the range `[2, 1000]`.
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- `1000 <= Node.val <= 1000`
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- The value of each node in the list is **unique**.
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- The `node` to be deleted is **in the list** and is **not a tail** node
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## 题目大意
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@ -37,4 +66,29 @@ Note:
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## 解题思路
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其实就是把后面的结点都覆盖上来即可。或者直接当前结点的值等于下一个结点,Next 指针指向下下个结点,这样做也可以,只不过中间有一个结点不被释放,内存消耗多一些。
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其实就是把后面的结点都覆盖上来即可。或者直接当前结点的值等于下一个结点,Next 指针指向下下个结点,这样做也可以,只不过中间有一个结点不被释放,内存消耗多一些。
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## 代码
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```go
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package leetcode
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import (
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"github.com/halfrost/LeetCode-Go/structures"
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)
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// ListNode define
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type ListNode = structures.ListNode
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/**
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* Definition for singly-linked list.
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* type ListNode struct {
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* Val int
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* Next *ListNode
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* }
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*/
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func deleteNode(node *ListNode) {
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node.Val = node.Next.Val
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node.Next = node.Next.Next
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}
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```
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