算法导论10.2链表

带哨兵的双向链表,代码中我使用了nullptr，所以需要编译器升级，我的编译器是gcc/g++ 4.7.0这是可以的，编译的时候加参数—std=c++0x

节点中还可能有卫星元素

/*
 * IA_10.2LinkedLists.cpp
 *
 *  Created on: Feb 12, 2015
 *      Author: sunyj
 */
#include <stdint.h>
#include <iostream>

class ListNode {
public:
    ListNode() : key(0), data(0), prev(nullptr), next(nullptr) { }
    ListNode(int64_t const k, int64_t d) : key(k), data(d), prev(nullptr), next(nullptr) { }

    int64_t key;
    int64_t data;
    ListNode* prev;
    ListNode* next;
};

// LIST-SEARCH(L, k)
// x = L.nil.next
// while ( x != L.nil and x.key != k)
//     x = x.next
// return x

// LIST-INSERT(L, x)
// x.next = L.nil.next
// L.nil.next.prev = x
// L.nil.next = x
// x.prev = L.nil

// LIST-DELETE(L, x)
// x.prev.next = x.next
// x.next.prev = x.prev

class LinkedList {
public:
    LinkedList() : nil(&m_nil)
    {
        nil->prev = nil;
        nil->next = nil;
        nil->data = -1; //
    }
    ListNode* search(int64_t const k)
    {
        ListNode* x = nil->next;
        while (x != nil && k != x->key)
        {
            x = x->next;
        }
        return x;
    }
    void insert(ListNode* x)
    {
        x->next         = nil->next;
        nil->next->prev = x;
        nil->next       = x;
        x->prev         = nil;
    }
    void del(ListNode* x)
    {
        x->prev->next = x->next;
        x->next->prev = x->prev;
    }
    void print()
    {
        ListNode* x = nil->next;
        while (nil != x)
        {
            std::cout << x->key << " ";
            x = x->next;
        }
        std::cout << std::endl;
    }
private:
    ListNode  m_nil; // empty list has noe node, pointer nill points to it.
    ListNode* nil;
};

int main()
{
    LinkedList list;
    ListNode node1(1, 100);
    ListNode node4(4, 400);
    ListNode node16(16, 1600);
    ListNode node9(9, 900);
    list.insert(&node1);
    list.insert(&node4);
    list.insert(&node16);
    list.insert(&node9);
    list.print();
    ListNode node25(25, 2500);
    list.insert(&node25);
    list.print();
    list.del(&node1);
    list.print();
    ListNode* tmp;
    tmp = list.search(9);
    list.del(tmp);
    list.print();
    return 0;
}