Title：
Write a function to delete a node (except the tail) in a singly linked list(单链表), 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（有效节点） of the linked list.
- Do not return anything from your function. ！！！！！
Analysis of Title：
We needn't to analysis the title but the Note end that Do not return anything from your function.
Test case：
[4,5,1,9]
5
Python：
class Solution(object):
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
Analysis of Code：
This is the content of the linked list.
The algorithm is:
1.[4,5,1,9]-->>[4,1,1,9]
2.[4,1,1,9]-->>[4,1,9]
Notice: node.next means ponit next node, just like 5->1, so node.next.next means 5->->9.
.next is a 'point', not a certain value, Originally I thought node.next is 1, so node.next = node.next.next means 5=9, it's wrong.