#Leetcode# 707. Design Linked List
https://leetcode.com/problems/design-linked-list/
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
valbefore 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
valto the last element of the linked list. - addAtIndex(index, val) : Add a node of value
valbefore theindex-th node in the linked list. Ifindexequals 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.
代码:
class MyLinkedList {
public:
/** Initialize your data structure here. */
MyLinkedList() {
head = NULL;
sz = 0;
}
/** Get the value of the index-th node in the linked list. If the index is invalid, return -1. */
int get(int index) {
if(index < 0 || index >= sz) return -1;
ListNode *cur = head;
for(int i = 0; i < index; i ++) cur = cur -> next;
return cur -> 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. */
void addAtHead(int val) {
ListNode *pre = new ListNode(val, head);
head = pre;
sz ++;
}
/** Append a node of value val to the last element of the linked list. */
void addAtTail(int val) {
ListNode *pre = head;
while(pre -> next) pre = pre -> next;
pre -> next = new ListNode(val, NULL);
sz ++;
}
/** 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. */
void addAtIndex(int index, int val) {
if(index < 0 || index > sz) return;
if(index == 0) {
addAtHead(val);
return;
}
ListNode *pre = head;
for(int i = 0; i < index - 1; i ++) pre = pre -> next;
ListNode *cur = new ListNode(val, pre -> next);
pre -> next = cur;
sz ++;
}
/** Delete the index-th node in the linked list, if the index is valid. */
void deleteAtIndex(int index) {
if(index < 0 || index >= sz) return;
if(index == 0) {
head = head -> next;
sz --;
return;
}
ListNode *pre = head;
for(int i = 0; i <index - 1; i ++) pre = pre -> next;
pre -> next = pre -> next -> next;
sz --;
}
private:
struct ListNode {
int val;
ListNode *next;
ListNode(int x, ListNode *n): val(x),next(n) {}
};
ListNode *head, *tail;
int sz;
};
/**
* Your MyLinkedList object will be instantiated and called as such:
* MyLinkedList obj = new MyLinkedList();
* int param_1 = obj.get(index);
* obj.addAtHead(val);
* obj.addAtTail(val);
* obj.addAtIndex(index,val);
* obj.deleteAtIndex(index);
*/
这个写完之后仿佛已经搞清楚了链表 但愿不是错觉叭
#Leetcode# 707. Design Linked List的更多相关文章
- 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_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 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】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 ...
- 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 ...
- [LeetCode] 641.Design Circular Deque 设计环形双向队列
Design your implementation of the circular double-ended queue (deque). Your implementation should su ...
- 【LeetCode】链表 linked list(共34题)
[2]Add Two Numbers (2018年11月30日,第一次review,ko) 两个链表,代表两个整数的逆序,返回一个链表,代表两个整数相加和的逆序. Example: Input: ( ...
- LeetCode解题报告:Linked List Cycle && Linked List Cycle II
LeetCode解题报告:Linked List Cycle && Linked List Cycle II 1题目 Linked List Cycle Given a linked ...
- [LeetCode] 92. Reverse Linked List II_Medium tag: Linked List
Reverse a linked list from position m to n. Do it in one-pass. Note: 1 ≤ m ≤ n ≤ length of list. Exa ...
随机推荐
- 北京Uber优步司机奖励政策(4月23日)
滴快车单单2.5倍,注册地址:http://www.udache.com/ 如何注册Uber司机(全国版最新最详细注册流程)/月入2万/不用抢单:http://www.cnblogs.com/mfry ...
- python基础学习2-easygui框架编程
#!/usr/bin/env python # -*- coding:utf-8 -*- import easygui as g #导入方式一 #导入方式2 #from easygui import ...
- pandas:对字符串类型做差分比较
1. 问题需求 某种行为最常发生时段.最少发生时段与X天前是否一致 需求变形:判断上下行数据是否一致 2. 预备知识 2.1 Series.ne(Series) 判断两个Series是否相等 impo ...
- 如何用python获取文件中的某一行——python小技巧
很多人有的时候只需要获取文章中的固定的一行,那么我知道这一行,我需要怎么样去获取呢 可能会有人说读取这一行,如果这一行是已什么开头的就读出来, 其实还有一种办法,就是我知道文件的路径.知道我要取的行数 ...
- python易错盲点排查之+=与+的区别分析以及一些赋值运算踩过的坑
问题1. int和list是不一样的 >>> a=1 >>> b=a >>> a+=1 >>> a,b (2, 1) >& ...
- python-GUI之tkinter的学习
最近看了哔哩哔哩的python的学习,直接看代码吧,以后会更新 先来个基础的 import tkinter as tk #导入包 app = tk.Tk() #抽象出一个GUI app.title(& ...
- SSO流程
SSO SSO又名单点登录,用户只需要登录一次就可以访问权限范围内的所有应用子系统.举个简单的例子,你在百度首页登录成功之后,你再访问百度百科.百度知道.百度贴吧等网站也会处于登录状态了,这就是一个单 ...
- python数据分析系列(1)
目录 python基础 python语言基础 Ipython的一些特性 Python语法基础 Python控制流 lambda表达式 Python的数据结构 元组 列表 字典 集合 列表.集合.字典推 ...
- 转载:GBDT算法梳理
学习内容: 前向分布算法 负梯度拟合 损失函数 回归 二分类,多分类 正则化 优缺点 sklearn参数 应用场景 转自:https://zhuanlan.zhihu.com/p/58105824 G ...
- CentOS 7 Docker基本特性
Docker是一个开源的应用容器引擎,开发人员可以非常容易地打包已经开发好的应用,同时将应用相关的依赖包也打包到这样一个可移植的容器中,然后发布到任意的Linux主机系统上.Docker是基于Linu ...