一道有趣的题，237. Delete Node in a Linked List

题目如下：

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.

这道题，只给了当前想要删除的节点，没有给这个节点前面的节点，导致删除时，无法使用先前节点指向之后节点的方法。

但是仔细分析这道题，确实能发现一些巧妙之处，题目虽然说：given only access to that node. 但是之前也说明了except the tail.

说明这个要被删除的节点，绝对不是最后一个，所以我们可以使用另一种方法：

把这个要被删除节点的下一个节点的值，赋值到这个节点，并且删除下一个节点，从结果上看，结果是一样的。

# 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