数据结构链表(二)

链表

双向链表的C++实现
单向链表的成环问题
环中元素个数
环开始的节点位置

双向链表的C++实现

#pragma once
typedef struct Node
{int value;Node* prev;Node* next;
}node;class DoubleLinklist
{private:int size;node* head;node* tail;
public:DoubleLinklist();~DoubleLinklist();//获取链表大小int GetSize();//判断是否为空bool IsEmpty();//尾插法插入元素节点void Insert(node* n);//头插法插入元素节点void HeadInsert(node* n);//指定位置后插入数据void PosInsert(int pos, node* n);//获取指定位置元素，并打印void GetPosElem(int pos);//搜索指定元素并打印位置void SearchElem(int elem);//删除指定元素void DelPosElem(int pos);//打印链表void ShowLinklist();//反转打印链表void ReverseShowLinklist();//清除所有数据void ClearLinklist();
};

#include"doublelinlist.h"
#include<iostream>
using std::cout;
using std::cin;
using std::endl;DoubleLinklist::DoubleLinklist()
{this->size = 0;this->head = new node;this->head->value = NULL;this->head->prev = this->head->next = NULL;this->tail = new node;this->tail->next = this->tail->prev = NULL;this->tail->value = NULL;
}DoubleLinklist::~DoubleLinklist()
{delete[]head;delete[]tail;
}int DoubleLinklist::GetSize()
{cout << "NOW LINKLIST SIZE == ";cout << this->size << endl;return this->size;
}bool DoubleLinklist::IsEmpty()
{if (this->size)return false;elsereturn true;
}void DoubleLinklist::Insert(node* n)
{if (n){node* temp = new node;temp->next = temp->prev = NULL;temp->value = n->value;if (this->IsEmpty()){this->head->next = temp;temp->next = this->tail;this->tail->prev = temp;temp->prev = this->head;}else{node* pnode = this->tail->prev;pnode->next = temp;temp->next = this->tail;this->tail->prev = temp;temp->prev = pnode;}this->size++;}else{cout << "UNDEFINE NODE\n";}}void DoubleLinklist::HeadInsert(node* n)
{if (n){node* temp = new node;temp->next = temp->prev = NULL;temp->value = n->value;if (this->IsEmpty()){this->head->next = temp;temp->next = this->tail;this->tail->prev = temp;temp->prev = this->head;}else{node* pnode = this->head->next;this->head->next = temp;temp->next = pnode;pnode->prev = temp;temp->prev = this->head;}this->size++;}else{cout << "UNDEFINED NODE\n";}
}void DoubleLinklist::PosInsert(int pos, node* n)//注意插入时前后节点的指针赋值顺序
{                                               //一般先将插入节点的前后指针next、prev赋值if (n)                                        //再改变前节点的next指针，后节点的prev指针{node* temp = new node;temp->next = temp->prev = NULL;temp->value = n->value;if (pos > this->size || pos < 0){cout << "NO SUCH POSITION, DEFULT INSERT IT IN THE END\n";this->Insert(n);}else if (pos == this->size)this->Insert(n);else if (pos == 0)this->HeadInsert(n);else{node* pnode;if (pos <= (this->size / 2)){pnode = this->head;for (int i = 0; i < pos; ++i){pnode = pnode->next;}temp->next = pnode->next;temp->prev = pnode;pnode->next->prev = temp;pnode->next = temp;}else{pnode = this->tail;for (int i = 0; i < (this->size - pos + 1); ++i){pnode = pnode->prev;}temp->next = pnode->next;temp->prev = pnode;pnode->next->prev = temp;pnode->next = temp;}}this->size++;}else{cout << "UNDEFINED NODE\n";}
}void DoubleLinklist::GetPosElem(int pos)
{if (pos <= 0 || pos > this->size)cout << "NO SUCH POSITION\n";else{node* pnode;if (pos <= (this->size / 2)){pnode = head;for (int i = 0; i < pos; ++i){pnode = pnode->next;}}else{pnode = tail;for (int i = 0; i < (this->size + 1 - pos); ++i){pnode = pnode->prev;}}cout << "POSITION [" << pos << "] == ";cout << pnode->value << endl;}
}void DoubleLinklist::SearchElem(int elem)
{if (this->IsEmpty())cout << "NO DATA!\n";else{node* temp = this->head;int flags = 0;for (int i = 0; i < this->size; ++i){temp = temp->next;if (temp->value == elem){flags++;cout << "FIND IT, POSITION == ";cout << i+1 << endl;}}if(!flags)cout << "THERE IS NO DATA YOU WANT\n";}
}void DoubleLinklist::DelPosElem(int pos)
{node* temp;if (pos <= 0 || pos > this->size)cout << "POSITION ERROR\n";else if (pos == 1){temp = this->head->next;head->next = temp->next;temp->next->prev = head;delete temp;}else if (pos == this->size){temp = this->tail->prev;tail->prev = temp->prev;temp->prev->next = this->tail;delete temp;}else{temp = this->head;for (int i = 0; i < pos; ++i){temp = temp->next;}temp->next->prev = temp->prev;temp->prev->next = temp->next;delete temp;}this->size--;
}void DoubleLinklist::ShowLinklist()
{if (this->IsEmpty())cout << "LINKLIST NULL\n";else{node* temp = this->head;for (int i = 0; i < this->size; ++i){temp = temp->next;cout << "[" << i + 1 << "] == " << temp->value << endl;}}
}void DoubleLinklist::ReverseShowLinklist()
{if (this->IsEmpty())cout << "LINKLIST NULL\n";else{node* temp = this->tail;for (int i = 0; i < this->size; ++i){temp = temp->prev;cout << "[" << i + 1 << "] == " << temp->value << endl;}}
}void DoubleLinklist::ClearLinklist()
{if (this->IsEmpty())cout << "LINKLIST NULL\n";else{node* temp = this->head->next;for (int i = 0; i < this->size; ++i){temp = temp->next;delete temp->prev;}}
}

#include"doublelinlist.h"
#include<iostream>
using std::cout;
using std::endl;int main()
{DoubleLinklist lk;node n[10];for (int i = 0; i < 10; ++i){n[i].value = 5 + i;n[i].next = n[i].prev = nullptr;}lk.Insert(&n[2]);lk.Insert(&n[4]);lk.Insert(&n[3]);lk.Insert(&n[1]);lk.Insert(&n[8]);lk.Insert(&n[5]);lk.Insert(&n[0]);lk.ShowLinklist();cout << endl;lk.ReverseShowLinklist();cout << "\n";lk.HeadInsert(&n[1]);lk.HeadInsert(&n[9]);lk.ShowLinklist();cout << endl;lk.ReverseShowLinklist();cout << "\n";lk.PosInsert(4, &n[6]);lk.PosInsert(6, &n[7]);lk.ShowLinklist();cout << "\n";lk.DelPosElem(3);lk.DelPosElem(4);lk.ShowLinklist();cout << "\n";lk.SearchElem(2);lk.SearchElem(17);lk.SearchElem(8);lk.ClearLinklist();return 0;
}

测试结果，基本没有错误出现

[1] == 7
[2] == 9
[3] == 8
[4] == 6
[5] == 13
[6] == 10
[7] == 5[1] == 5
[2] == 10
[3] == 13
[4] == 6
[5] == 8
[6] == 9
[7] == 7[1] == 14
[2] == 6
[3] == 7
[4] == 9
[5] == 8
[6] == 6
[7] == 13
[8] == 10
[9] == 5[1] == 5
[2] == 10
[3] == 13
[4] == 6
[5] == 8
[6] == 9
[7] == 7
[8] == 6
[9] == 14[1] == 14
[2] == 6
[3] == 7
[4] == 9
[5] == 11
[6] == 8
[7] == 12
[8] == 6
[9] == 13
[10] == 10
[11] == 5[1] == 14
[2] == 6
[3] == 9
[4] == 8
[5] == 12
[6] == 6
[7] == 13
[8] == 10
[9] == 5THERE IS NO DATA YOU WANT
THERE IS NO DATA YOU WANT
FIND IT, POSITION == 4

单向链表的成环问题

通过数学归纳法简单推断，即可得出使用快慢指针，快指针2格一跳，慢指针1格一跳，快慢指针终会在环里相遇，并且在相遇之时，慢指针还未走满一个环。

//承接上一章中单向链表类，存在一个头节点
bool Linklist::LinklistIsRing()
{if (this->size == 0)return false;node* fast = this->head->next;node* slow = this->head->next;while (fast != NULL && fast->next != NULL){fast = fast->next->next;slow = slow->next;if (fast == slow)return true;}return false;
}

环中元素个数

int Linklist::RingCount()
{if (this->size == 0)return -1;node* fast = this->head->next;node* slow = this->head->next;int count = 0;while (fast != NULL && fast->next != NULL){fast = fast->next->next;slow = slow->next;if (fast == slow)break;}if (fast == NULL)//追加判断不为环的情况下直接退出return -1;do{slow = slow->next;count++;} while (slow == fast);return count;
}

环开始的节点位置

node* Linklist::GetRingStart()
{if (this->size == 0)return NULL;node* fast = this->head->next;node* slow = this->head->next;int count = 0;while (fast != NULL && fast->next != NULL){fast = fast->next->next;slow = slow->next;if (fast == slow)break;}if (fast == NULL)return NULL;slow = this->head->next;//慢指针回归到头，快指针更改速度为一个，下次相遇即为环的开始节点while (slow != fast){slow = slow->next;fast = fast->next;}return slow;
}

数据结构链表(二)相关推荐

SDUT _2117 数据结构实验之链表二：逆序建立链表
点击打开链接数据结构实验之链表二:逆序建立链表 Time Limit: 1000MS Memory Limit: 65536KB Submit Statistic Discuss Problem D ...
数据结构（二）——单链表的头插与尾插
一.引言上篇文章我们知道了顺序表的缺点: 插入和删除操作需要移动大量元素. 数组的大小不好确定. 存储分配需要一整段连续的存储空间,不够灵活,造成很多碎片(空闲的空间得不到利用). 所以我们就引入了 ...
数据结构（二）--队列
数据结构(二)–队列文章目录数据结构(二)--队列介绍代码实现数组方式实现普通队列环形队列链式实现单向队列介绍队列是一种特殊的线性表,特殊之处在于它只允许在表的前端(front) ...
剑指offer(C++)-JZ76：删除链表中重复的结点(数据结构-链表)
作者:翟天保Steven 版权声明:著作权归作者所有,商业转载请联系作者获得授权,非商业转载请注明出处题目描述: 在一个排序的链表中,存在重复的结点,请删除该链表中重复的结点,重复的结点不保留,返回 ...
剑指offer(C++)-JZ35：复杂链表的复制(数据结构-链表)
作者:翟天保Steven 版权声明:著作权归作者所有,商业转载请联系作者获得授权,非商业转载请注明出处题目描述: 输入一个复杂链表(每个节点中有节点值,以及两个指针,一个指向下一个节点,另一个特殊指 ...
剑指offer(C++)-JZ23：链表中环的入口结点(数据结构-链表)
作者:翟天保Steven 版权声明:著作权归作者所有,商业转载请联系作者获得授权,非商业转载请注明出处题目描述: 给一个长度为n链表,若其中包含环,请找出该链表的环的入口结点,否则,返回null. ...
剑指offer(C++)-JZ52：两个链表的第一个公共结点(数据结构-链表)
作者:翟天保Steven 版权声明:著作权归作者所有,商业转载请联系作者获得授权,非商业转载请注明出处题目描述: 输入两个无环的单向链表,找出它们的第一个公共结点,如果没有公共节点则返回空.(注意因 ...
剑指offer(C++)-JZ25：合并两个排序的链表(数据结构-链表)
作者:翟天保Steven 版权声明:著作权归作者所有,商业转载请联系作者获得授权,非商业转载请注明出处题目描述: 输入两个递增的链表,单个链表的长度为n,合并这两个链表并使新链表中的节点仍然是递增排 ...
剑指offer(C++)-JZ24：反转链表(数据结构-链表)
作者:翟天保Steven 版权声明:著作权归作者所有,商业转载请联系作者获得授权,非商业转载请注明出处题目描述: 给定一个单链表的头结点pHead,长度为n,反转该链表后,返回新链表的表头. 数据范 ...

数据结构链表(二)

链表

双向链表的C++实现

单向链表的成环问题

环中元素个数

环开始的节点位置

数据结构链表(二)相关推荐

最新文章

热门文章

数据结构 链表(二)

链表

双向链表的C++实现

单向链表的成环问题

环中元素个数

环开始的节点位置

数据结构 链表(二)相关推荐

最新文章

热门文章

数据结构链表(二)

数据结构链表(二)相关推荐