leetcode: 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:

img

Example 1:

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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:

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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.

题目描述有点问题,题目的原意是给一个node,直接在该node存在的list中删除这个node,因为不知道父节点的地址,所以只需要将node的val改成node.next的val,然后把node.next设置成node.next.next就可以了。

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# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None

class :
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