237: Delete Node in a Linked List

题目

Write a function to delete a node (except the tail) in a singly linked list, given only access to that node.

Supposed the linked list is 1 -> 2 -> 3 -> 4 and you are given the third node with value 3, the linked list should become 1 -> 2 -> 4 after calling your function.

题目大意：
编写一个函数删除单链表中（除末尾节点外）的一个节点，只提供待删除节点。

假如链表是1 -> 2 -> 3 -> 4 给你第3个节点，值为3，则调用你的函数后链表为1 -> 2 -> 4 

解答

• 开始的时候写反了,写成

node.next.val = node.val
node.next.next = node.next

产生错误

Last executed input:
[0,1]
node at index 0 (node.val = 0)

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

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