(C++)有非循环双链表L，需按照按值查找Locate（L，x）的频度排序，查找频率越高，越靠近头结点。 L的 结点中包含一个频度域，每次查找后，频度域+1，并调整结点在L中的位置，然后返回指向该结点

解题思路：

1. 构建并初始化非循环双向链表；
2. 数据域储存值和频度，让频度开始都为0
3. 按值函数查找后所在结点频度加1。
4. 所得结点和前面结点频度比较，如果频度大于等于前面结点频度，那么交换两个结点，否则不交换，位置不变。
5. 当交换时，循环比较，直到所换位置在头节点后面或者和小于前面结点。

关键代码：

1. 双链表的尾插法：

void begin(Node* head, Date a[], int n)

{

    Node* rear = head;



    for (int i = 0; i < n; ++i)

    {

        Node* s = new Node;

        s->next = nullptr;

        s->prior = nullptr;

        s->date = a[i].date;

        s->free = a[i].free;



        rear->next = s;

        s->prior = rear;

        rear = s;

    }

}

1. 结点和前面结点的交换：

void begin(Node* head, Date a[], int n)

{

    Node* rear = head;



    for (int i = 0; i < n; ++i)

    {

        Node* s = new Node;

        s->next = nullptr;

        s->prior = nullptr;

        s->date = a[i].date;

        s->free = a[i].free;

        rear->next = s;

        s->prior = rear;

        rear = s;

    }

}

1. 按值查找：

Node *Locate(Node *L,int x)

    {

    Node *p= L->next;

        while (p != nullptr && p->date != x)

        p = p->next;

        if( p->date!=x) return 0;

    p->free=p->free+1;

    while(1)

    {

        if (p->prior != L &&p->free >= p->prior->free)

            change(p);

            else break;

    }

    return p;

    }

完整代码：

#include<iostream>
using namespace std;

struct Node {
    int date;
    int free;
    Node* next = nullptr;
    Node* prior = nullptr;
};
struct Date
{
    int date;
    int free;
};

void destroy(Node* head)
{
    Node* p = head;
    while (p)
    {
        Node* q = p;
        p = p->next;
        delete q;
    }
}
void show(Node* t2)
{
    if (t2 != nullptr)
    {
        cout << "初始数据为：" << endl;
        t2 = t2->next;
        while (t2 != nullptr)
        {
            cout << "   " << t2->date << "(" << t2->free << ")";
            t2 = t2->next;
        }
        cout << endl;
    }
    else
        cout << "表中无数据" << endl;
}
void begin(Node* head, Date a[], int n)
{
    Node* rear = head;

    for (int i = 0; i < n; ++i)
    {
        Node* s = new Node;
        s->next = nullptr;
        s->prior = nullptr;
        s->date = a[i].date;
        s->free = a[i].free;

        rear->next = s;
        s->prior = rear;
        rear = s;
    }
}
void change(Node* L)
{

    Node* p = L; Node* q = p->prior;
    if (p->next != nullptr)
    {
        q->prior->next = p;
        p->prior = q->prior;
        q->next = p->next;
        p->next->prior = q;
        p->next = q;
        q->prior = p;
    }
    else
    {
        p->prior = q->prior;
        p->next = q;
        q->prior->next = p;
        q->prior = p;
        q->next=nullptr;

    }

}
Node *Locate(Node *L,int x)
    {
    Node *p= L->next;

        while (p != nullptr && p->date != x)
        p = p->next;
        if( p->date!=x) return 0;
    p->free=p->free+1;

    while(1)
    {
        if (p->prior != L &&p->free >= p->prior->free)
            change(p);
            else break;

    }
    return p;
    }
int main()
{
    Date a[10];
    for (int i = 0; i < 10; ++i)
    {
        a[i].date = i + 1;
        a[i].free = 0;
    }


    Node* head = new Node;
    head->next = head;
    head->prior = head;

    begin(head,a,10);
    show(head);
    Node* p = head->next->next;
    Node* q = head->next;
    while (1)
    {
        int x;
        cout << "请输入需要查询的值：" << endl;
        cin >> x;
     Node *xx=Locate(head, x);
        show(head);
       cout <<"所查询的指针的地址为：" << xx << endl<<endl;
    }
    destroy(head);
    return 0;


}

结果截图：