707. Design Linked List
1. 原始题目
Design your implementation of the linked list. You can choose to use the singly linked list or the doubly linked list. A node in a singly linked list should have two attributes: val
and next
. val
is the value of the current node, and next
is a pointer/reference to the next node. If you want to use the doubly linked list, you will need one more attribute prev
to indicate the previous node in the linked list. Assume all nodes in the linked list are 0-indexed.
Implement these functions in your linked list class:
- get(index) : Get the value of the
index
-th node in the linked list. If the index is invalid, return-1
. - addAtHead(val) : Add a node of value
val
before the first element of the linked list. After the insertion, the new node will be the first node of the linked list. - addAtTail(val) : Append a node of value
val
to the last element of the linked list. - addAtIndex(index, val) : Add a node of value
val
before theindex
-th node in the linked list. Ifindex
equals to the length of linked list, the node will be appended to the end of linked list. If index is greater than the length, the node will not be inserted. - deleteAtIndex(index) : Delete the
index
-th node in the linked list, if the index is valid.
Example:
MyLinkedList linkedList = new MyLinkedList();
linkedList.addAtHead(1);
linkedList.addAtTail(3);
linkedList.addAtIndex(1, 2); // linked list becomes 1->2->3
linkedList.get(1); // returns 2
linkedList.deleteAtIndex(1); // now the linked list is 1->3
linkedList.get(1); // returns 3
Note:
- All values will be in the range of
[1, 1000]
. - The number of operations will be in the range of
[1, 1000]
. - Please do not use the built-in LinkedList library.
2. 题目理解
设计链表的实现。您可以选择使用单链表或双链表。单链表中的节点应该具有两个属性:val
和 next
。val
是当前节点的值,next
是指向下一个节点的指针/引用。如果要使用双向链表,则还需要一个属性 prev
以指示链表中的上一个节点。假设链表中的所有节点都是 0-index 的。
在链表类中实现这些功能:
- get(index):获取链表中第
index
个节点的值。如果索引无效,则返回-1
。 - addAtHead(val):在链表的第一个元素之前添加一个值为
val
的节点。插入后,新节点将成为链表的第一个节点。 - addAtTail(val):将值为
val
的节点追加到链表的最后一个元素。 - addAtIndex(index,val):在链表中的第
index
个节点之前添加值为val
的节点。如果index
等于链表的长度,则该节点将附加到链表的末尾。如果index
大于链表长度,则不会插入节点。 - deleteAtIndex(index):如果索引
index
有效,则删除链表中的第index
个节点。
示例:
MyLinkedList linkedList = new MyLinkedList();
linkedList.addAtHead(1);
linkedList.addAtTail(3);
linkedList.addAtIndex(1,2); //链表变为1-> 2-> 3
linkedList.get(1); //返回2
linkedList.deleteAtIndex(1); //现在链表是1-> 3
linkedList.get(1); //返回3
提示:
- 所有值都在
[1, 1000]
之内。 - 操作次数将在
[1, 1000]
之内。 - 请不要使用内置的 LinkedList 库。
注意·的点:空链表,索引范围为0~超出链表长度~我改了两次才通过。
3. 解题
class ListNode: # 定义结点类型
def __init__(self, x):
self.val = x
self.next = None class MyLinkedList: def __init__(self):
"""
Initialize your data structure here.
"""
self.head = None # 初始化一个头结点为空 def get(self, index: int) -> int: # 返回第index个结点,若index不合法则返回-1
"""
Get the value of the index-th node in the linked list. If the index is invalid, return -1.
"""
if self.head == None: # 空链表对于所有index都不合法
return -1
p = self.head
for i in range(index):
if not p.next:
return -1
p = p.next
return p.val def addAtHead(self, val: int) -> None: # 在头部插入一个结点
"""
Add a node of value val before the first element of the linked list. After the insertion, the new node will be the first node of the linked list.
"""
if self.head == None: # 对空链表直接将该结点赋予头结点即可
self.head = ListNode(val)
else:
new_node = ListNode(val)
new_node.next = self.head
self.head = new_node # 更新头结点 def addAtTail(self, val: int) -> None: # 在尾部插入一个结点
"""
Append a node of value val to the last element of the linked list.
"""
if self.head == None: # 对空链表直接将该结点赋予头结点即可
self.head = ListNode(val)
p = self.head
while(p.next):
p = p.next
p.next = ListNode(val) def addAtIndex(self, index: int, val: int) -> None: # 在第index位置插入结点
"""
Add a node of value val before the index-th node in the linked list. If index equals to the length of linked list, the node will be appended to the end of linked list. If index is greater than the length, the node will not be inserted.
"""
p = self.head
new_node = ListNode(val)
if not self.head: # 若为空链表,除非index=0,才将其作为头结点,否则一切index都不合法
if index==0:
new_node.next = self.head
self.head = new_node
return None
for i in range(index-1):
if not p.next:
return None
p = p.next
new_node.next = p.next
p.next = new_node def deleteAtIndex(self, index: int) -> None: # 删除index位置的结点
"""
Delete the index-th node in the linked list, if the index is valid.
"""
if not self.head: # 空链表退出
return None
if index == 0: # 头结点单独考虑,直接将头结点赋予下一个结点即可
self.head = self.head.next
return None
p = self.head
for i in range(index-1):
if not p.next:
return None
p = p.next
if p.next:
p.next = p.next.next
4. 验证
验证之前可以写一个print函数打印当前链表情况:写到类里面去
def printlist(self):
p = self.head
while(p):
print(p.val,end=' ')
p = p.next
print('\n')
测试:
linkedList = MyLinkedList() linkedList.addAtHead(5)
linkedList.printlist() linkedList.addAtHead(2)
linkedList.printlist() linkedList.deleteAtIndex(1)
linkedList.printlist() linkedList.addAtIndex(1,9)
linkedList.printlist() linkedList.addAtHead(4)
linkedList.printlist() linkedList.addAtHead(9)
linkedList.printlist() linkedList.addAtHead(8)
linkedList.printlist() print(linkedList.get(3) ) linkedList.addAtTail(1)
linkedList.printlist() linkedList.addAtIndex(3,6)
linkedList.printlist() linkedList.addAtHead(3)
linkedList.printlist()
5
2 5
2
2 9
4 2 9
9 4 2 9
8 9 4 2 9
2
8 9 4 2 9 1
8 9 4 6 2 9 1
3 8 9 4 6 2 9 1
707. Design Linked List的更多相关文章
- 【Leetcode_easy】707. Design Linked List
problem 707. Design Linked List 参考 1. Leetcode_easy_707. Design Linked List; 完
- #Leetcode# 707. Design Linked List
https://leetcode.com/problems/design-linked-list/ Design your implementation of the linked list. You ...
- 【LeetCode】707. Design Linked List 解题报告(Python)
作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 日期 题目地址:https://leetcode.c ...
- LeetCode 707. Design Linked List (设计链表)
题目标签:Linked List 题目让我们自己设计一个 linked list,可以是单向和双向的.这里选的是单向,题目并不是很难,但要考虑到所有的情况,具体看code. Java Solution ...
- [LeetCode] Design Linked List 设计链表
Design your implementation of the linked list. You can choose to use the singly linked list or the d ...
- [Swift]LeetCode707. 设计链表 | Design Linked List
Design your implementation of the linked list. You can choose to use the singly linked list or the d ...
- 【LeetCode】Design Linked List(设计链表)
这道题是LeetCode里的第707到题.这是在学习链表时碰见的. 题目要求: 设计链表的实现.您可以选择使用单链表或双链表.单链表中的节点应该具有两个属性:val 和 next.val 是当前节点的 ...
- LeetCode707:设计链表 Design Linked List
爱写bug (ID:iCodeBugs) 设计链表的实现.您可以选择使用单链表或双链表.单链表中的节点应该具有两个属性:val 和 next.val 是当前节点的值,next 是指向下一个节点的指针/ ...
- Design Linked List
Design your implementation of the linked list. You can choose to use the singly linked list or the d ...
随机推荐
- gometalinter代码质量检查分析工具(golang)
GitHub地址:https://github.com/alecthomas/gometalinter gometalinter安装和使用 1.安装 go get github.com/alectho ...
- 2017-12-14python全栈9期第一天第八节之循环语句while
12,while. while 条件: 循环体 无限循环. 终止循环:1,改变条件,使其不成立. 2,break continue
- 获取APP的元素信息和Activity
一.获取元素信息 (1)第一种方法是,在windows命令行中输入uiautomatorviewer.bat(前提是已配置好Android SDK环境),按回车键:等待几秒后会打开UI Automat ...
- 简:Spring中Bean的生命周期及代码示例
(重要:spring bean的生命周期. spring的bean周期,装配.看过spring 源码吗?(把容器启动过程说了一遍,xml解析,bean装载,bean缓存等)) 完整的生命周期概述(牢记 ...
- ZOJ - 3261 Connections in Galaxy War(并查集删边)
https://cn.vjudge.net/problem/ZOJ-3261 题意 银河系各大星球之间有不同的能量值, 并且他们之间互相有通道连接起来,可以用来传递信息,这样一旦有星球被怪兽攻击,便可 ...
- PHP7 学习笔记(十一)使用phpstudy快速配置一个虚拟主机
说明:为了windows本地开发php方便,这里推荐使用PHP集成环境phpstudy. 目的:使用域名访问项目(tinywan.test) 1.官网:http://www.phpstudy.net ...
- vue watch bug记录
watch中,写箭头函数,获取不到正确的this 换成function,正确取到this
- Devexpress dll搜集
Devexpress一部分在全局dll中,需要分析缺哪些dll,有两种方式1.打包,安装时会自动提示 2.使用自带分析工具Assembly deployment tool
- extjs.net 按钮执行并显示Mask代码
<ext:Button ID="ButtonTest" runat="server" Width="65" Text="同 ...
- Android adb shell 常用命令
Ø 简介 adb 命令是 Android SDK 中自带的一个调试工具,可以调试电脑中的手机或者模拟器,使用 adb 命令前需要将 sdk 中的 platform-tools 目录添加到环境变量中. ...