5-237. Delete Node in a Linked List

题目描述：

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.

代码实现（大神版）：

 1 # Definition for singly-linked list.
 2 # class ListNode:
 3 #     def __init__(self, x):
 4 #         self.val = x
 5 #         self.next = None
 6 
 7 class Solution:
 8     def deleteNode(self, node):
 9         """
10         :type node: ListNode
11         :rtype: void Do not return anything, modify node in-place instead.
12         """
13         node.val = node.next.val
14         node.next = node.next.next

分析：

关于用Python实现链表的知识详见：https://www.cnblogs.com/kumata/p/9147077.html

题目的要求是给定一个节点（该节点不在链表的尾部），将该节点删除掉。

实现思路非常简单：只需要用下一个节点来覆盖要删除的节点即可，也就是让当前节点的内容等于下一个节点的内容，同时让当前节点的指针指向下一个节点的指针即可。

如果给出的不是一个节点，而是给出节点中的内容，让删除这个节点，则一种代码实现方式如下：（个人版）

 1 class ListNode:
 2     def __init__(self, x):
 3         self.val = x
 4         self.next = None
 5 
 6 class Solution:
 7     def deleteNode(self, node):
 8         """
 9         :type node: ListNode
10         :rtype: void Do not return anything, modify node in-place instead.
11         """
12         a = ListNode(9)
13         b = ListNode(1)
14         c = ListNode(5)
15         d = ListNode(4)
16 
17         b.next = a
18         c.next = b
19         d.next = c
20         head = d
21 
22         fpointer, bpointer = head, head
23         while True:
24             if fpointer.val == node:
25                 if fpointer == head:
26                     head.val = fpointer.next.val
27                     head.next = fpointer.next.next
28                 elif fpointer.next == None:
29                     bpointer.next = None
30                 else:
31                     bpointer.next = fpointer.next
32                     bpointer.next.val = fpointer.next.val
33                 break
34             else:
35                 bpointer = fpointer
36                 fpointer = fpointer.next
37         
38         pointer = head
39         aa = [0] * 3
40         i = 0
41         while i < 3:
42             aa[i] = pointer.val
43             pointer = pointer.next
44             i += 1
45         print(aa)
46 
47 if __name__ == '__main__':
48     Solution().deleteNode(9)