刪除一個鏈表節點(Leetcode)

problem address

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.

參數只有一個待刪除的節點node,比如刪除節點5, 可以用1的值賦值給5，然後5指向9來達到刪除5的目的

# 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.
        """
        nextnode = node.next
        after_next_node = nextnode.next
        node.val = nextnode.val
        node.next = after_next_node