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)
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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
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Leetcode 237 Delete node in a Linked List
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