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.
这题有点旱路不走走水路的意思。
一开始我也觉得这题出错了， 我们无法拿到要删除节点的上一个节点。但是按照官方 solution 的说法， 我们可以“假装”删掉当前节点， 怎么假装呢， 就是我们把当前节点的 Val 改成它的 Next 的 Val， 然后把 Next 删掉, 这样的效果就跟删掉当前节点一样了。
这次用golang实现的， rust实现可能只能用unsafe了， leetcode压根就没给准备模板
func deleteNode(node *ListNode) {
node.Val = node.Next.Val
node.Next = node.Next.Next
}