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LeetCode---Delete Node in a Linked List
阅读量:606 次
发布时间:2019-03-12

本文共 1254 字,大约阅读时间需要 4 分钟。

237. Delete Node in a Linked List

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 = 5Output: [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 = 1Output: [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.

请编写一个函数,使其可以删除某个链表中给定的(非末尾)节点,你将只被给定要求被删除的节点。

思路:现在只给了当前节点,那么只能将后一节点的值赋给当前节点,将后一节点删掉,则相当于删掉了“当前节点”。

# class ListNode(object):#     def __init__(self, x):#         self.val = x#         self.next = Noneclass 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

 

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