Delete Node in a Linked List

https://leetcode.com/explore/featured/card/june-leetcoding-challenge/539/week-1-june-1st-june-7th/3348/

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:

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.

删除给定的链表结点。由于只给了待删除的结点，不可能是真正意义上的『删除』，是要实现删除的效果，也就是把下一个值赋给当前结点，并把此时多余的下个结点删除。

# 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