第一篇终结Linked List(一)终结Linked List(二)主要讲了单链表的基础知识,接下来的第二篇主要讲一些比较经典的问题。

一、Count()

给一个单链表和一个整数,返回这个整数在链表中出现了多少次。

/*
Given a list and an int, return the number of times
that int ocucurs in the list.
*/
int Count(struct node* head,int searchFor)
{
int cnt = 0;
struct node* cur = head; while (cur != NULL)
{
if (cur->data == searchFor)
cnt++;
cur = cur->next;
} return cnt;
}

也可以用for循环实现。

二、GetNth()

给一个单链表和一个index,返回index位置上的数值,类似array[index]操作。

/*
Given a list and an index, return the data in the nth
node of the list. The nodes are numbered from 0.
Assert fails if the index is invalid (outside 0..length - 1).
*/
int GetNth(struct node* head,int index)
{
int len = 0;
struct node* cur = head; while (cur)
{
if (len == index)
{
return cur->data;
}
cur = cur->next;
len++;
} assert(0); //如果走到这一行,表达式的值为假,断言失败
}

三、DeleteList()

给一个单链表,删除所有节点,使headNULL

删除链表{1,2,3}的示意图:

void DeleteList(struct node** headRef)
{
struct node* cur = *headRef; //deref headRef to get the real head while (*headRef)
{
cur = *headRef;
*headRef = cur->next;
free(cur);
}
}

四、Pop()

给一个链表,删掉头节点,返回头节点的数据。

内存示意图:

/*
The opposite of Push().Takes a non-empty list and
remove the front node, and returns the data which was in that node.
*/
int pop(struct node** headRef)
{
assert(*headRef != NULL);
int ans = (*headRef)->data; //pull out the data before the node is deleted struct node* cur = *headRef;
*headRef = (*headRef)->next; //unlink the head node for the caller
free(cur); //free the head node return ans;
}

五、InsertNth()

可以在[0,length]的任意位置插入指定元素。

/*
A more general version of Push().
Given a list, an index 'n' in the range 0..length,
and a data element, add a new node to the list so that
it has the given index.
*/
void InsertNth(struct node** headRef,int index,int data)
{
//position 0 is a special case
if (index == 0)
{
Push(headRef, data);
}
else
{
int cnt = 0;
struct node* cur = *headRef; while (cnt < index - 1)
{
assert(cur != NULL); //if this fails, the index was too big
cur = cur->next;
cnt++;
} assert(cur != NULL); //tricky:you have to check one last time Push(&(cur->next), data);
}
}

这段代码坑有点多,可以通过画图或者单步跟踪的方法调试。

InsertNthTest()可以用来测试:

void InsertNthTest()
{
struct node* head = NULL; //start with the empty list InsertNth(&head, 0, 13); //{13}
InsertNth(&head, 1, 42); //{13,42}
InsertNth(&head, 1, 5); //{13,5,42}
}

六、SortedInsert()

给定一个有序链表和一个节点,将该节点插入到合适的位置。

共有三种方法:

1、Uses special case code for the head end

void SortedInsert(struct node** headRef,struct node* newNode)
{
//Special case for the head end
if (newNode->data <= (*headRef)->data || *headRef == NULL)
{
newNode->next = *headRef;
*headRef = newNode;
}
else
{
//Locate the node before the point of insertion
struct node* cur = *headRef;
while (cur->next && cur->next->data < newNode->data)
{
cur = cur->next;
}
newNode->next = cur->next;
cur->next = newNode;
}
}

2、Dummy node strategy for the head end

dummy node这种方法一般不需要处理特殊情况。

void SortedInsert2(struct node** headRef,struct node* newNode)
{
struct node dummy;
struct node* cur = &dummy;
dummy.next = *headRef; while (cur->next && newNode->data >= cur->next->data)
{
cur = cur->next;
}
newNode->next = cur->next;
cur->next = newNode; *headRef = dummy.next; //头指针永远指向dummy.next
}

3、Local references strategy for the head end

void SortedInsert3(struct node** headRef,struct node* newNode)
{
struct node** curRef = headRef; while (*curRef && (*curRef)->data <= newNode->data)
{
curRef = &((*curRef)->next);
} newNode->next = *curRef; //Bug:(*curRef)->next is incorrect *curRef = newNode;
}

Linked List-3的更多相关文章

  1. [LeetCode] Linked List Random Node 链表随机节点

    Given a singly linked list, return a random node's value from the linked list. Each node must have t ...

  2. [LeetCode] Plus One Linked List 链表加一运算

    Given a non-negative number represented as a singly linked list of digits, plus one to the number. T ...

  3. [LeetCode] Odd Even Linked List 奇偶链表

    Given a singly linked list, group all odd nodes together followed by the even nodes. Please note her ...

  4. [LeetCode] Delete Node in a Linked List 删除链表的节点

    Write a function to delete a node (except the tail) in a singly linked list, given only access to th ...

  5. [LeetCode] Palindrome Linked List 回文链表

    Given a singly linked list, determine if it is a palindrome. Follow up: Could you do it in O(n) time ...

  6. [LeetCode] Reverse Linked List 倒置链表

    Reverse a singly linked list. click to show more hints. Hint: A linked list can be reversed either i ...

  7. [LeetCode] Remove Linked List Elements 移除链表元素

    Remove all elements from a linked list of integers that have value val. Example Given: 1 --> 2 -- ...

  8. [LeetCode] Intersection of Two Linked Lists 求两个链表的交点

    Write a program to find the node at which the intersection of two singly linked lists begins. For ex ...

  9. [LeetCode] Linked List Cycle II 单链表中的环之二

    Given a linked list, return the node where the cycle begins. If there is no cycle, return null. Foll ...

  10. [LeetCode] Linked List Cycle 单链表中的环

    Given a linked list, determine if it has a cycle in it. Follow up: Can you solve it without using ex ...

随机推荐

  1. Tkinter 控件

    文章更新于:2020-02-19 待翻译跟进 In this part of the Tkinter tutorial, we will cover some basic Tkinter widget ...

  2. Kafka,RocketMQ,RabbitMQ部署与使用体验

    前言 近期在研究各种消息队列方案,为了有一个直观的使用体验,我把Kafka,RocketMQ,RabbitMQ各自部署了一遍,并使用了最基本的生产与消费消息功能.在部署过程中也遇到一些问题,特此记录. ...

  3. 20 java 基础回顾--中阶引入

    一.数据类型 基本数据类型(共:四类八种) 整数 byte short int long 浮点 float double 字符 char 布尔 boolean 引用数据类型(new的数据) Stude ...

  4. CH5701 开车旅行(倍增dp+set)

    传送门 解题思路: 一道比较有趣的题,解题工作主要分为两块: ①找出k(k=0表示小A先走,k=1表示小B先走,下面同理)从点i出发下一个到达的点to[k][i]; 一开始偷懒用了vector(偷懒一 ...

  5. break与continue对比

    - break 用来终止循环 - continue 用来跳出当前循环,继续下次循环 // 求1到100之间所有不能被3整除的整数的第一个大于2000的和 var sum = 0; for(var i= ...

  6. AJ学IOS 之微博项目实战(9)微博模型之时间相关重要操作,判断刚刚,昨天,今年等等

    AJ分享,必须精品 一:效果 二:实现代码 /** 1.今年 1> 今天 * 1分内: 刚刚 * 1分~59分内:xx分钟前 * 大于60分钟:xx小时前 2> 昨天 * 昨天 xx:xx ...

  7. 常用Linux命令的基本使用

    01.常用Linux命令的基本使用 序号 命令 对应英文 作用 01 ls list 查看当前文件夹下的内容 02 pwd print work directory 查看当前所在文件夹 03 cd [ ...

  8. Python爬虫某招聘网站的岗位信息

    前言 文的文字及图片来源于网络,仅供学习.交流使用,不具有任何商业用途,版权归原作者所有,如有问题请及时联系我们以作处理. 作者:阿尔法游戏 PS:如有需要Python学习资料的小伙伴可以加点击下方链 ...

  9. 无序map 记录一下

    unordered_map<int ,int >mp; unordered_map是基于hash表实现的,查找元素的复杂度可以达到o(1),查找n个元素,复杂度为o(n). map是基于红 ...

  10. First Training

    B B - Local Extrema CodeForces - 888A You are given an array a. Some element of this array ai is a l ...