python - 以简单的方式从单链表中删除节点

Question

我正在通过一个简单的leetcode问题学习LinkedList Delete Node in a Linked List - LeetCode

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.

官方解决方案:

将要删除的节点的值替换为其后节点中的值，然后删除其后的节点。

# 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.
            """
            node.val = node.next.val 
            node.next = node.next.next 

我觉得麻烦，所以改成:

class Solution:
    def deleteNode(self, node):
        node= node.next

以简单的方式将下一个节点分配给当前节点

不幸的是，它报告了错误的答案。

我不明白为什么它不起作用。

逻辑没有错误:当前节点被下一个节点遮蔽，

Answer 1

这就是我们想要做的，删除节点:

前 -> 节点 -> 后 => 前 -> 后

因此，您应该做的是将 pre.next 从 node 更改为 node.next。像:

pre.next = node.next

如果您执行node = node.next，它只会更改node引用。

对原来的ListNode没有任何影响。

但是在这种情况下我们无法获取 pre，因此我们必须将 node 值更改为 post，并删除 发布。

希望对您有帮助，如有疑问请评论。 :)