数据结构与算法编程题11

已知两个链表A和B分别表示两个集合，其元素递增排列。
请设计算法求出A与B的交集，并存放于A链表中。
a: 1, 2, 2, 4, 5, 7, 8, 9, 10
b: 1, 2, 3, 6, 7, 8

#include 
using namespace std;

typedef int Elemtype;
#define ERROR 0;
#define OK    1;

typedef struct LNode
{
	Elemtype data;      //结点保存的数据
	struct LNode* next; //结构体指针
}LNode, * LinkList;

/*单链表初始化*/
bool Init_LinkList(LinkList& L)
{
	L = (LinkList)malloc(sizeof(LNode));  //新建头结点
	if (L == NULL)
	{
		return ERROR;
	}
	L->data = 0;
	L->next = NULL;
	return OK;
}

/*单链表头插法*/
bool LinkList_head_instert(LinkList& L)
{
	int x = 0;
	LNode* p = NULL;
	while (cin >> x)
	{
		p = (LinkList)malloc(sizeof(LNode));
		if (p != NULL)  //防止分配地址失败
		{
			p->data = x;
			p->next = L->next;
			L->next = p;
			if (cin.get() == '\n') break;  //检测换行符
		}
		else
		{
			exit(0);
			cout << "内存分配失败" << endl;
		}
	}
	return OK;
}

/*单链表尾插法*/
bool LinkList_tail_instert(LinkList& L)
{
	int x = 0;
	LNode* p = NULL;
	LNode* r = NULL;
	r = L;
	while (cin >> x)
	{
		p = (LinkList)malloc(sizeof(LNode));
		if (p != NULL)  //防止分配地址失败
		{
			p->data = x;
			p->next = NULL;
			r->next = p;
			r = p;
			if (cin.get() == '\n') break;  //检测换行符
		}
		else
		{
			exit(0);
			cout << "内存分配失败" << endl;
		}
	}
	return OK;
}

/*单链表遍历*/
bool LinkList_All_value(LinkList L)
{
	if (L->next == NULL)
	{
		cout << "链表为空" << endl;
		return ERROR;
	}
	LNode* s = NULL;
	s = L->next;
	while (s != NULL)
	{
		cout << s->data << "   ";
		s = s->next;
	}
	cout << endl;
	free(s);
	return OK;
}

/*单链表长度*/
int LinkList_length(LinkList L)
{
	int count = 0;
	LNode* s = NULL;
	s = L->next;
	while (s != NULL)
	{
		count++;
		s = s->next;
	}
	return count;
}

/*清空单链表*/
void Clear_LinkList(LinkList& L)
{
	LNode* p = L->next;
	LNode* q = NULL;
	while (p != NULL)
	{
		q = p->next;
		free(p);
		p = q;
	}
	L->next = NULL;
}

/*销毁单链表*/
void Destory_LinkList(LinkList& L)
{
	LNode* p = NULL;
	LNode* q = NULL;
	p = L;
	while (p != NULL)
	{
		q = p->next;
		free(p);
		p = q;
	}
	L = NULL;
}

bool jiaoji(LinkList& La, LinkList& Lb)
{
	LNode* pa = NULL;
	LNode* pb = NULL;
	LNode* pc = NULL;
	LNode* q = NULL;
	pa = La->next;
	pb = Lb->next;
	pc = La;
	La->next = NULL;
	if (pa == NULL && pb == NULL)
	{
		cout << "两个单链表为空!!!" << endl;
		return ERROR;
	}
	while (pa != NULL && pb != NULL)
	{
		if (pa->data == pb->data)
		{
			pc->next = pa;
			pc = pa;
			pa = pa->next;
			q = pb;
			pb = pb->next;
			delete q;//或者用free(q);
		}
		else if (pa->data > pb->data)
		{
			q = pb;
			pb = pb->next;
			delete q;
		}
		else //pa->data < pb->data
		{
			q = pa;
			pa = pa->next;
			delete q;
		}
	}
	while (pa != NULL)
	{
		q = pa;
		pa = pa->next;
		delete q;
	}
	while (pb != NULL)
	{
		q = pb;
		pb = pb->next;
		delete q;
	}
	pc->next = NULL;
	delete Lb;
	return OK;
}

/*已知两个链表A和B分别表示两个集合，其元素递增排列。
请设计算法求出A与B的交集，并存放于A链表中。*/
//a: 1, 2, 2, 4, 5, 7, 8, 9, 10
//b: 1, 2, 3, 6, 7, 8

int	main(void)
{
	LinkList a = NULL;
	Init_LinkList(a);
	LinkList_tail_instert(a);//1 2 2 4 5 7 8 9 10
	LinkList_All_value(a);
	LinkList b = NULL;
	Init_LinkList(b);
	LinkList_tail_instert(b);//1 2 3 6 7 8
	LinkList_All_value(b);
	jiaoji(a, b);
	LinkList_All_value(a);//打印两个单链表的交集
	return 0;
}