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:

4 -> 5 -> 1 -> 9

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.

# Definition for singly-linked list.class ListNode:    def __init__(self, x):        self.val = x        self.next = Noneclass Solution:    def deleteNode(self, node):        """        :type node: ListNode        :rtype: void Do not return anything, modify node in-place instead.        """        if node is None or node.next is None:            return        node.val = node.next.val        node.next = node.next.next        return

题目已经给了提示，把下个节点的值放到node里，删掉下个节点就可以了。