Leetcode 237 Delete node in a Linked List

Delete node in a Linked List"/>

Leetcode 237 Delete node in a Linked List

Leetcode 237 Delete node in a Linked List

问题描述：

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:

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.

来源：力扣（LeetCode）
链接：
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

问题解析

这道题要求我们从一个单链表中删除一个指定的结点，并且限制我们无法访问该链表的所有结点，只能访问待删除的结点。

从链表中删除一个结点，我们以往通常的做法是获取该结点的上一个结点p，利用p.next = p.next.next来删除该结点。而这道题目中我们无法获取到待删除节点的上一个结点，这就要求我们用另一个思路去删除：将待删除结点的值替换为下一个结点的值，然后删除下一个结点（因待删除结点不是末尾结点，所有这一操作是可以实现的），从而实现删除指定结点。

利用几张图来说明，比如我们需要删除值为3的结点：

• 以往通常的思路

• 这道题的思路

• 启示：一个操作的实现往往有多种方法，我们不能仅局限于一种，使自己陷入思维定式。

代码实现

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = Noneclass Solution:def deleteNode(self, node):""":type node: ListNode:rtype: void Do not return anything, modify node in-place instead."""node.val = node.next.valnode.next = node.next.next