237. Delete Node in a Linked List

题目

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: [1]

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

题目大意

删除给点结点。没有给链表的头结点。

解题思路

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。

代码

package leetcode

import (
	"github.com/halfrost/LeetCode-Go/structures"
)

// ListNode define
type ListNode = structures.ListNode

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func deleteNode(node *ListNode) {
	node.Val = node.Next.Val
	node.Next = node.Next.Next
}

Uh oh!