算法题：求解两个链表的交加

/*
已知集合A和B的元素分别用不含头结点的单链表存储，
函数difference()用于求解集合A与B的差集，
并将结果保存在集合A的单链表中。例如，
若集合A = { 5, 10, 20, 15, 25, 30 }，
集合B = { 5, 15, 35, 25 }，完成计算
后A = { 10, 20, 30 }。
*/
#include <iostream>
using namespace std;
struct Node
{
    int data;
    Node *next;
    Node(int d = int()):data(d),next(NULL){}
};

class List
{
public:
    List(int a[],int n):head(NULL)
    {
        int i = 0;
        Node *p = NULL;
        for (; i<n; i++)
        {
            Node *s = new Node(a[i]);
            if (head == NULL)
            {
                head = s;
                p = head;
                s = p;
            }
            else
            {
                s->next = p->next;
                p->next = s;
                p = s;
            }
        }
    }
    void difference(const List &list)//求交集。
    {
        Node *pA = head;
        Node *pB = list.head;
        Node *savP=NULL;
        if (pA == NULL || pB == NULL)
        {
            head = NULL;
            return;
        }
        Node *temp=NULL;
        while (pA != NULL && pB!=NULL)  
        {
            while (pA->data < pB->data)
            {
                pA=pA->next;
                if (pA == NULL)
                    return;
            }
            while (pA->data > pB->data)
            {
                pB=pB->next;
                if (pB == NULL)
                    return;
            }
            if (pA->data == pB->data)
            {
                Node *m = pA->next;
                if (savP == NULL)
                {
                    savP = pA;
                    savP->next = NULL;
                }
                else
                {
                    temp = savP;
                    while (temp->next != NULL)
                    {
                        temp = temp->next;
                    }
                    pA->next = NULL;
                    temp->next = pA;
                }
                pA=m;
                pB=pB->next;
            }
        }
        head = savP;

    }
    void Printf()
    {
        Node *p = head;
        while (p != NULL)
        {
            cout << p->data << "---->";
            p = p->next;
        }
        cout << endl;
    }
private:
    Node *head;
};


int main()
{
    int a[] = {1,2,3,4,5,6,7,8,9};
    int b[] = {4,5,6,7,8,9,10};
    List list1(a, sizeof(a) / sizeof(int));
    List list2(b, sizeof(b) / sizeof(int));
    list1.difference(list2);
    list1.Printf();
    list2.Printf();
    return 0;
}