LeetCode＃237. Delete nodes in the linked list

Please write a function , be used for Delete a specific node in the single linked list . When designing functions, you need to pay attention to , You can't access the head node of the linked list head , Direct access only The node to be deleted .

The topic data ensures that the nodes to be deleted Not the end node .

Example 1：

Input ：head = [4,5,1,9], node = 5
Output ：[4,1,9]
explain ： Specifies that the value in the linked list is  5  Second node of , So after calling your function , The list should be 4 -> 1 -> 9
Example 2：

Input ：head = [4,5,1,9], node = 1
Output ：[4,5,9]
explain ： Specifies that the value in the linked list is  1  The third node of , So after calling your function , The list should be 4 -> 5 -> 9
Example 3：

Input ：head = [1,2,3,4], node = 3
Output ：[1,2,4]
Example 4：

Input ：head = [0,1], node = 0
Output ：[1]
Example 5：

Input ：head = [-3,5,-99], node = -3
Output ：[5,-99]
 

Tips ：

The number range of nodes in the linked list is [2, 1000]
-1000 <= Node.val <= 1000
The value of each node in the linked list is unique
Nodes to be deleted node yes A valid node in the linked list , And Not the end node

source ： Power button （LeetCode）
link ： Power button

This problem can be solved as long as we delete the specified node , Because the title has stated that this node is not the tail node , We can put node.val Locate the node.next.val, meanwhile node.next=node.next.next that will do .

# 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