数据结构开发(9):循环链表与双向链表

0.目录

1.循环链表的实现

2.双向链表的实现

3.小结

1.循环链表的实现

什么是循环链表?

  • 概念上
    1. 任意数据元素都有一个前驱和一个后继
    2. 所有的数据元素的关系构成一个逻辑上的环
  • 实现上
    1. 循环链表是一种特殊的单链表
    2. 尾结点的指针域保存了首结点的地址

循环链表的逻辑构成:

循环链表的继承层次结构:

循环链表的实现思路:

  • 通过模板定义CircleList类,继承自LinkList
  • 定义内部函数 last_to_first(),用于将单链表首尾相连
  • 特殊处理:首元素的插入操作和删除操作
  • 重新实现:清空操作和遍历操作

循环链表的实现要点:

  • 插入位置为 0 时:
    1. 头结点和尾结点均指向新结点
    2. 新结点成为首结点插入链表
  • 删除位置为 0 时:
    1. 头结点和尾结点指向位置为 1 的结点
    2. 安全销毁首结点

实现循环链表(CircleList.h):
需要将父类LinkList.h中相关的函数声明为虚函数

#ifndef LINKLIST_H
#define LINKLIST_H

#include "List.h"
#include "Exception.h"

namespace StLib
{

template <typename T>
class LinkList : public List<T>
{
protected:
    struct Node : public Object
    {
        T value;
        Node* next;
    };

    mutable struct : public Object
    {
        char reserved[sizeof(T)];
        Node* next;
    } m_header;

    int m_length;
    int m_step;
    Node* m_current;

    Node* position(int i) const
    {
        Node* ret = reinterpret_cast<Node*>(&m_header);

        for(int p=0; p<i; p++)
        {
            ret = ret->next;
        }

        return ret;
    }

    virtual Node* create()
    {
        return new Node();
    }

    virtual void destroy(Node* pn)
    {
        delete pn;
    }

public:
    LinkList()
    {
        m_header.next = NULL;
        m_length = 0;
        m_step = 1;
        m_current = NULL;
    }

    bool insert(const T& e)
    {
        return insert(m_length, e);
    }

    bool insert(int i, const T& e)
    {
        bool ret = ((0 <= i) && (i <= m_length));

        if( ret )
        {
            Node* node = create();

            if( node != NULL )
            {
                Node* current = position(i);

                node->value = e;
                node->next = current->next;
                current->next = node;

                m_length++;
            }
            else
            {
                THROW_EXCEPTION(NoEnoughMemoryException, "No memory to insert new element ...");
            }
        }

        return ret;
    }

    bool remove(int i)
    {
        bool ret = ((0 <= i) && (i < m_length));

        if( ret )
        {
            Node* current = position(i);
            Node* toDel = current->next;

            if( m_current == toDel )
            {
                m_current = toDel->next;
            }

            current->next = toDel->next;

            m_length--;

            destroy(toDel);
        }

        return ret;
    }

    bool set(int i, const T& e)
    {
        bool ret = ((0 <= i) && (i < m_length));

        if( ret )
        {
            position(i)->next->value = e;
        }

        return ret;
    }

    virtual T get(int i) const
    {
        T ret;

        if( get(i, ret) )
        {
            return ret;
        }
        else
        {
            THROW_EXCEPTION(IndexOutOfBoundsException, "Invalid parameter i to get element ...");
        }

        return ret;
    }

    bool get(int i, T& e) const
    {
        bool ret = ((0 <= i) && (i < m_length));

        if( ret )
        {
            e = position(i)->next->value;
        }

        return ret;
    }

    int find(const T& e) const
    {
        int ret = -1;
        int i = 0;
        Node* node = m_header.next;

        while ( node )
        {
            if( node->value == e )
            {
                ret = i;
                break;
            }
            else
            {
                node = node->next;
                i++;
            }
        }

        return ret;
    }

    int length() const
    {
        return m_length;
    }

    void clear()
    {
        while ( m_header.next )
        {
            Node* toDel = m_header.next;

            m_header.next = toDel->next;

            m_length--;

            destroy(toDel);
        }
    }

    virtual bool move(int i, int step = 1)
    {
        bool ret = (0 <= i) && (i < m_length) && (step > 0);

        if( ret )
        {
            m_current = position(i)->next;
            m_step = step;
        }

        return ret;
    }

    virtual bool end()
    {
        return (m_current == NULL);
    }

    virtual T current()
    {
        if( !end() )
        {
            return m_current->value;
        }
        else
        {
            THROW_EXCEPTION(InvalidOperationException, "No value at current position ...");
        }
    }

    virtual bool next()
    {
        int i = 0;

        while( (i < m_step) && !end() )
        {
            m_current = m_current->next;
            i++;
        }

        return (i == m_step);
    }

    ~LinkList()
    {
        clear();
    }
};

}

#endif // LINKLIST_H

实现CircleList.h

#ifndef CIRCLELIST_H
#define CIRCLELIST_H

#include "LinkList.h"

namespace StLib
{

template <typename T>
class CircleList : public LinkList<T>
{
protected:
    typedef typename LinkList<T>::Node Node;

    int mod(int i) const
    {
        return (this->m_length == 0) ? 0 : (i % this->m_length);
    }

    Node* last() const
    {
        return this->position(this->m_length-1)->next;
    }

    void last_to_first() const
    {
        last()->next = this->m_header.next;
    }
public:
    bool insert(const T& e)
    {
        return insert(this->m_length, e);
    }

    bool insert(int i, const T& e)
    {
        bool ret;

        i = i % (this->m_length + 1);

        ret = LinkList::insert(i, e);

        if( ret && (i == 0) )
        {
            last_to_first();
        }

        return ret;
    }

    bool remove(int i)
    {
        bool ret;

        i = mod(i);

        if( i == 0 )
        {
            Node* toDel = this->m_header.next;

            if( toDel != NULL )
            {
                this->m_header.next = toDel->next;
                this->m_length--;

                if( this->m_length > 0 )
                {
                    last_to_first();

                    if( this->m_current == toDel )
                    {
                        this->m_current = toDel->next;
                    }
                }
                else
                {
                    this->m_header.next = NULL;
                    this->m_current = NULL;
                }

                this->destroy(toDel);
            }
            else
            {
                ret = false;
            }
        }
        else
        {
            ret = LinkList<T>::remove(i);
        }

        return ret;
    }

    bool set(int i, const T& e)
    {
        return LinkList<T>::set(mod(i), e);
    }

    T get(int i) const
    {
        return LinkList<T>::get(mod(i));
    }

    T get(int i, const T& e) const
    {
        return LinkList<T>::get(mod(i), e);
    }

    int find(const T& e) const
    {
        int ret = -1;
        Node* slider = this->m_header.next;

        for(int i=0; i<this->m_length; i++)
        {
            if( slider->value == e )
            {
                ret = i;
                break;
            }

            slider = slider->next;
        }

        return ret;
    }

    void clear()
    {
        while( this->m_length > 1 )
        {
            remove(1);
        }

        if( this->m_length == 1 )
        {
            Node* toDel = this->m_header.next;

            this->m_header.next = NULL;
            this->m_length = 0;
            this->m_current = NULL;

            this->destroy(toDel);
        }
    }

    bool move(int i, int step)
    {
        return LinkList<T>::move(mod(i), step);
    }

    bool end()
    {
        return (this->m_length == 0) || (this->m_current == NULL);
    }

    ~CircleList()
    {
        clear();
    }
};

}

#endif // CIRCLELIST_H

循环链表的应用——约瑟夫环问题:

已知 n 个人( 以编号 0,1,2,3,... ,n-1 分别表示 )围坐在一张圆桌周围。从编号为 k 的人开始报数,数到 m 的那个人出列;他的下一个人又从 1 开始报数,数到 m 的那个人又出列;依此规律重复下去,直到圆桌周围的人全部出列。

小故事:

在罗马人占领乔塔帕特后,39个犹太人与 Josephus 及他的朋友躲到一个洞中,39个犹太人决定宁愿死也不要被敌人抓到,于是决定了一个自杀方式,41个人排成一个圆圈,由第1个人开始报数,每报数到第3人该人就必须自杀,然后再由下一个重新报数,直到所有人都自杀身亡为止。然而 Josephus 和他的朋友并不想遵从。那么,一开始要站在什么地方才能避免被处决?

main.cpp解决约瑟夫环问题,测试CircleList.h:

#include <iostream>
#include "CircleList.h"

using namespace std;
using namespace StLib;

void josephus(int n, int s, int m)
{
    CircleList<int> c1;

    for(int i=1; i<=n; i++)
    {
        c1.insert(i);
    }

    c1.move(s-1, m-1);

    while( c1.length() > 0 )
    {
        c1.next();

        cout << c1.current() << ", ";

        c1.remove(c1.find(c1.current()));
    }
    cout << endl;
}

int main()
{
    josephus(41, 1, 3);

    return 0;
}

运行结果为:

3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 1, 5, 10, 14, 19, 23, 28, 32, 37, 41, 7, 13, 20, 26, 34, 40, 8, 17, 29, 38, 11, 25, 2, 22, 4, 35, 16, 31, 

2.双向链表的实现

单链表的另一个缺陷:

  • 单向性
    1. 只能从头结点开始高效访问链表中的数据元素
  • 缺陷
    1. 如果需要逆向访问单链表中的数据元素将极其低效

新的线性表

  • 设计思路:
    1. 在“单链表”的结点中增加一个指针 pre,用于指向当前结点的前驱结点。

双向链表的继承层次结构:

DualLinkList 的定义:

实现DualLinkList.h:

#ifndef DUALLINKLIST_H
#define DUALLINKLIST_H

#include "List.h"
#include "Exception.h"

namespace StLib
{

template <typename T>
class DualLinkList : public List<T>
{
protected:
    struct Node : public Object
    {
        T value;
        Node* next;
        Node* pre;
    };

    mutable struct : public Object
    {
        char reserved[sizeof(T)];
        Node* next;
        Node* pre;
    } m_header;

    int m_length;
    int m_step;
    Node* m_current;

    Node* position(int i) const
    {
        Node* ret = reinterpret_cast<Node*>(&m_header);

        for(int p=0; p<i; p++)
        {
            ret = ret->next;
        }

        return ret;
    }

    virtual Node* create()
    {
        return new Node();
    }

    virtual void destroy(Node* pn)
    {
        delete pn;
    }

public:
    DualLinkList()
    {
        m_header.next = NULL;
        m_header.pre = NULL;
        m_length = 0;
        m_step = 1;
        m_current = NULL;
    }

    bool insert(const T& e)
    {
        return insert(m_length, e);
    }

    bool insert(int i, const T& e)
    {
        bool ret = ((0 <= i) && (i <= m_length));

        if( ret )
        {
            Node* node = create();

            if( node != NULL )
            {
                Node* current = position(i);
                Node* next = current->next;

                node->value = e;

                node->next = next;
                current->next = node;

                if( current != reinterpret_cast<Node*>(&m_header) )
                {
                    node->pre = current;
                }
                else
                {
                    node->pre = NULL;
                }

                if( next != NULL )
                {
                    next->pre = node;
                }

                m_length++;
            }
            else
            {
                THROW_EXCEPTION(NoEnoughMemoryException, "No memory to insert new element ...");
            }
        }

        return ret;
    }

    bool remove(int i)
    {
        bool ret = ((0 <= i) && (i < m_length));

        if( ret )
        {
            Node* current = position(i);
            Node* toDel = current->next;
            Node* next = toDel->next;

            if( m_current == toDel )
            {
                m_current = next;
            }

            current->next = next;

            if( next != NULL )
            {
                next->pre = toDel->pre;
            }

            m_length--;

            destroy(toDel);
        }

        return ret;
    }

    bool set(int i, const T& e)
    {
        bool ret = ((0 <= i) && (i < m_length));

        if( ret )
        {
            position(i)->next->value = e;
        }

        return ret;
    }

    virtual T get(int i) const
    {
        T ret;

        if( get(i, ret) )
        {
            return ret;
        }
        else
        {
            THROW_EXCEPTION(IndexOutOfBoundsException, "Invalid parameter i to get element ...");
        }

        return ret;
    }

    bool get(int i, T& e) const
    {
        bool ret = ((0 <= i) && (i < m_length));

        if( ret )
        {
            e = position(i)->next->value;
        }

        return ret;
    }

    int find(const T& e) const
    {
        int ret = -1;
        int i = 0;
        Node* node = m_header.next;

        while ( node )
        {
            if( node->value == e )
            {
                ret = i;
                break;
            }
            else
            {
                node = node->next;
                i++;
            }
        }

        return ret;
    }

    int length() const
    {
        return m_length;
    }

    void clear()
    {
        while ( m_length > 0 )
        {
            remove(0);
        }
    }

    virtual bool move(int i, int step = 1)
    {
        bool ret = (0 <= i) && (i < m_length) && (step > 0);

        if( ret )
        {
            m_current = position(i)->next;
            m_step = step;
        }

        return ret;
    }

    virtual bool end()
    {
        return (m_current == NULL);
    }

    virtual T current()
    {
        if( !end() )
        {
            return m_current->value;
        }
        else
        {
            THROW_EXCEPTION(InvalidOperationException, "No value at current position ...");
        }
    }

    virtual bool next()
    {
        int i = 0;

        while( (i < m_step) && !end() )
        {
            m_current = m_current->next;
            i++;
        }

        return (i == m_step);
    }

    virtual bool pre()
    {
        int i = 0;

        while( (i < m_step) && !end() )
        {
            m_current = m_current->pre;
            i++;
        }

        return (i == m_step);
    }

    ~DualLinkList()
    {
        clear();
    }
};

}

#endif // DUALLINKLIST_H

main.cpp测试

#include <iostream>
#include "DualLinkList.h"

using namespace std;
using namespace StLib;

int main()
{
    DualLinkList<int> d1;

    for(int i=0; i<5; i++)
    {
        d1.insert(0, i);
        d1.insert(0, 5);
    }

    cout << "begin" << endl;

    d1.move(d1.length()-1);

    while( !d1.end() )
    {
        if( d1.current() == 5 )
        {
            cout << d1.current() << endl;

            d1.remove(d1.find(d1.current()));
        }
        else
        {
            d1.pre();
        }
    }

    cout << "end" << endl;

    for(d1.move(d1.length()-1); !d1.end(); d1.pre())
    {
        cout << d1.current() << endl;
    }

    return 0;
}

运行结果为:

begin
5
5
5
5
5
end
0
1
2
3
4

深度思考——开放性问题:

  • DualLinkListLinkList 中存在很多完全一样的代码,如何进行重构降低代码的冗余性?冗余代码的出现是否意味着 DualLinkListLinkList 之间应该是继承关系

扩展练习——双向链表的子类:

3.小结

  • 循环链表是一种特殊的单链表
  • 尾结点的指针域保存了首结点的地址
  • 特殊处理首元素的插入操作和删除操作
  • 重新实现清空操作遍历操作
  • 双向链表是为了弥补单链表的缺陷而重新设计的
  • 在概念上,双向链表不是单链表,没有继承关系
  • 双向链表中的游标能够直接访问当前结点的前驱和后继
  • 双向链表是线性表概念的最终实现( 更贴近理论上的线性表 )
posted @ 2018-12-16 15:30  PyLearn  阅读(337)  评论(0编辑  收藏  举报