blocks|key|908965|text|显而易见的解决办法是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|908966|def+delete_by_index(node,+index):
++++for+_+in+range(index):
++++++++prev_node,+node+=+node,+node.next
++++prev_node.next+=+node.next|code-block|syntax|javascript|908967|但是，这将无法删除第一个节点。为了做到这一点，您需要让它返回新的列表头，这个列表头通常是旧的列表头，但是在删除头的情况下，它将是旧的第二个节点。所以：|908968|def+delete_by_index(node,+index):
++++if+not+index:
++++++++return+node.next
++++head+=+node
++++for+_+in+range(index):
++++++++prev_node,+node+=+node,+node.next
++++prev_node.next+=+node.next
++++return+head|908969|如何简化这一点应该是显而易见的，您应该这样做。|908970|另一种选择是剔除“搜索”和“删除”部分。编写一个nth函数(这应该很简单)，然后您可以这样做：|offset|length|style|CODE|908971|def+delete_by_index(node,+index):
++++if+not+index:
++++++++return+node.next
++++prev_node+=+nth(node,+index-1)
++++prev_node.next+=+prev_node.next.next
++++return+node|908972|如果您了解递归函数，您可能还想知道如何递归编写delete_by_index或nth+(这是最容易编写的递归函数之一)。|908973|您还可能想要捕获错误，例如删除10节点列表的第15节点，并使其更好。找出在这种情况下会出现什么错误(或者，如果您找不出，只需运行它并查看)，然后try/except，然后引发一个IndexError。|908974|在进行此操作时，您可能需要添加一个delete_by_data(node,+data)函数，或者一个delete_by_identity(node,+child_node)，以供进一步练习。|908975|entityMap^0|0|0|0|0|0|O|3|0|0|N|F|13|3|0|20|3|24|6|2H|A|0|H|Q|1E|10|0^^$0|@$1|2|3|4|5|6|7|12|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|14|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|15|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|16|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|17|8|@$O|18|P|19|Q|R]]|9|@]|A|$]]|$1|S|3|T|5|D|7|1A|8|@]|9|@]|A|$E|F]]|$1|U|3|V|5|6|7|1B|8|@$O|1C|P|1D|Q|R]|$O|1E|P|1F|Q|R]]|9|@]|A|$]]|$1|W|3|X|5|6|7|1G|8|@$O|1H|P|1I|Q|R]|$O|1J|P|1K|Q|R]|$O|1L|P|1M|Q|R]]|9|@]|A|$]]|$1|Y|3|Z|5|6|7|1N|8|@$O|1O|P|1P|Q|R]|$O|1Q|P|1R|Q|R]]|9|@]|A|$]]|$1|10|3|-4|5|6|7|1S|8|@]|9|@]|A|$]]]|11|$]]

The obvious solution is this:

<pre><code>def delete_by_index(node, index):
 for _ in range(index):
 prev_node, node = node, node.next
 prev_node.next = node.next
</code></pre>

However, this won't be able to delete the first node. In order to do that, you need to make it return the new list head, which will normally be the old list head, but will instead be the old second node in the case where you deleted the head. So:

<pre><code>def delete_by_index(node, index):
 if not index:
 return node.next
 head = node
 for _ in range(index):
 prev_node, node = node, node.next
 prev_node.next = node.next
 return head
</code></pre>

It should be obvious how to simplify this, and you should do so.

Another option is to factor out the "searching" and "deleting" parts. Write an <code>nth</code> function (this should be easy), then you can do this:

<pre><code>def delete_by_index(node, index):
 if not index:
 return node.next
 prev_node = nth(node, index-1)
 prev_node.next = prev_node.next.next
 return node
</code></pre>

If you know about recursive functions, you might also want to figure out how to write either <code>delete_by_index</code> or <code>nth</code> recursively (it's one of the easiest recursive functions to write).

You may also want to trap the error caused by deleting, say, the 15th node of a 10-node list and make it something nicer. Figure out what error you would get in that case (or, if you can't figure it out, just run it and see), then <code>try</code>/<code>except</code> that, and raise an <code>IndexError</code> instead.

While you're at it, you may want to add a <code>delete_by_data(node, data)</code> function and maybe a <code>delete_by_identity(node, child_node)</code> for further practice.

blocks|key|1784162|text|v->+w->+x+->+y>+z，如果我们想删除x，那么新的链表是v->+w->+y->+z，并且我们只能访问x。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1784163|下一个节点到x的位置，即节点Y的位置|1784164|next_node+=+x.next|1784165|将Y交换为X，然后删除Y+`x.data+=+next_node.data|1784166|x.next+=+next_data.next|1784167|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|M|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|N|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|O|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|P|8|@]|9|@]|A|$]]|$1|J|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|K|$]]

v-&gt; w-&gt; x-&gt;y&gt; z and if we want to delete x so that new Linked List is v -&gt; w-&gt; y-&gt; z and we have access to only x
<h1>position of the next node to x i.e. position of the Node Y</h1>
next_node = x.next
<h1>swapping the Y to X and then deleting Y `x.data = next_node.data</h1>
x.next = next_data.next

blocks|key|908999|text|在单链接列表中使用python删除节点。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|909000|def删除(self，data)：|909001|++++if+self.head.data==data:

++++++++temp=self.head.next
++++++++del+self.head
++++++++self.head=temp

++++else:


++++++p=self.head
++++++while+p.next.data!=data:
++++++++++p=p.next

++++++temp=p.next.next
++++++del+p.next
++++++p.next=temp|code-block|syntax|javascript|909002|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Deletion of a node using python in single linked list.

def delete(self,data):

<pre><code> if self.head.data==data:

 temp=self.head.next
 del self.head
 self.head=temp

 else:


 p=self.head
 while p.next.data!=data:
 p=p.next

 temp=p.next.next
 del p.next
 p.next=temp
</code></pre>

blocks|key|905800|text|按值删除：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|905801|def+delete(head,+val):

+++++if+head+is+None:
+++++++return+None


+++++prev+=+head
+++++temp+=+head.next

+++++while+temp+is+not+None:

+++++++++if+temp.val+==+val:
+++++++++++++prev.next+=+temp.next
+++++++++++++return+val

+++++++++else:
+++++++++++++prev+=+temp
+++++++++++++temp+=+temp.next

+++++return+None|code-block|syntax|javascript|905802|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Delete by value:

<pre><code>def delete(head, val):

 if head is None:
 return None


 prev = head
 temp = head.next

 while temp is not None:

 if temp.val == val:
 prev.next = temp.next
 return val

 else:
 prev = temp
 temp = temp.next

 return None
</code></pre>

I am trying to learn linked list in Using python,

Could someone please guide me how to delete a particular give node from a linked list?

<pre><code>#!/usr/bin/python 

class Node(object):
 def __init__(self, data=None, next=None):
 self.data = data
 self.next = next
 def __str__(self):
 return str(self.data)

def print_list(node):
 while node:
 print node,
 node = node.next
 print

def delete_node(node, node_to_remove):
 if first_node == None:
 return
 pass

# way of creating linked list
def create_linked_list1(n):
 linked_list = Node(1)
 head = linked_list
 for i in range(1, n):
 head.next = Node(i)
 head = head.next
 return linked_list

node1 = create_linked_list1(10)

print_list(node1)
</code></pre>

How to delete a given node from a linked list using Python

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

文章

问答

视频

学习中心

腾讯云实验室

直播

竞赛

腾讯云代码分析专区

腾讯iOA零信任安全管理系统专区

腾讯云架构师技术同盟交流圈

腾讯云数据库专区

腾讯云顾问专区

腾讯云原生专区

腾讯混元专区

腾讯云TCE专区

腾讯云Lighthouse专区

腾讯云HAI专区

腾讯云Edgeone专区

腾讯云存储专区

腾讯云智能专区

腾讯轻联专区 

腾讯云开发专区

TAPD专区

腾讯轻量云游戏服专区

腾讯云最具价值专家

腾讯云架构师技术同盟

腾讯云创作之星

腾讯云开发者先锋

腾讯云代码助手

云原生构建

TAPD 敏捷项目管理

Cloud Studio

SDK中心

API中心

命令行工具

涵盖代码开发、场景应用、自动测试全流程，助你从零构建专属AI助手

一站式MCP教程库，解锁AI应用新玩法

我正在尝试学习使用python的链接列表，请有人指导我如何从链接列表中删除一个特定的给定节点。#!/usr/bin/python                                                                                                ...

问如何使用Python从链接列表中删除给定节点
EN

Stack Overflow用户

社区

活动

圈层

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问如何使用Python从链接列表中删除给定节点EN