DHU programming exercise

6 Find the intersection of sequences （ Linked list ）

1. problem
Use the single linked list programming of the leading node ：

There are two sequences , They represent two sets .

Find their intersection and output .
2. Enter description
The first line is the input sequence A Information about ：

The first integer n(0<=n<=100), Expressing common ownership n Elements , After that n It's an integer , Express n Data of elements

The first line is the input sequence B Information about ：

The first integer n(0<=n<=100), Expressing common ownership n Elements , After that n It's an integer , Express n Data of elements

The output shows that
Output a sequence of elements that intersect , See example for output format .

If the intersection is empty , The output “head–>tail”

The order of elements in the intersection , According to its sequence A Order in .

such as ：

Sequence ：

A：5 3 2 7

B：1 3 5 8

Then the intersection is 5 and 3, Because in the sequence A in ,5 stay 3 In front of , So in the intersection 5 Also in the 3 In front of .
3. Example
Input
4 5 3 2 7
4 1 3 5 8
Output
head–>5–>3–>tail
4. Code

#include<iostream>
using namespace std;
struct node

{
    
	int  data;
	struct node *next;
};

struct node* create(int n ,int num[])

{
    

	int i;

	struct node *current;// Current insertion position pointer 

	struct node *tail;// Tail pointer 

	struct node *head;// The head pointer 

	head = (struct node*)malloc(sizeof(struct node));// Head node 

	head->next = NULL;

	current = head;// The current position pointer points to the header node 

	tail = head;// The tail pointer also points to the head node 

	for (i = 0; i < n; i++)

	{
    
			struct node *p;

			p = (struct node*)malloc(sizeof(struct node));

			p->data = num[i];

			p->next = current->next;

			current->next = p;

			current = p;

			if (current->next == NULL) tail = current; // The current position is at the back , It needs to be modified tail The pointer 
	}

	return head;

}


int main()

{
    
	int a[101], b[101] ,n,i,m;
	struct node *L1, *L2;
	cin >> n;
	for (i = 0; i < n; i++)
		cin >> a[i];
	cin >> m;
	for (i = 0; i < m; i++)
		cin >> b[i];
	L1 = create(n, a);
	L2 = create(m, b);


	// seek L1,l2 Intersection 
	struct node *p, *q;
	p = L1->next;
	q = L2->next;
	cout << "head";
	while (p)
	{
    
		while (q)
		{
    
			if (p->data == q->data)
			{
    
				cout <<"-->"<< p->data;
				break;// about L1 Every element in , All in L2 Find out if there are the same elements in , If you have any , Explain the intersection , Exit search directly , seek L1 The next element of is in L2 The corresponding point in 
			}
			q = q->next;
		}
		p = p->next;// Description traversal L1 The next element in 
		q = L2->next;// The key point is , For each L1 The elements in , Every time I have to go through it from the beginning L2 Look for the same elements 
	}
	cout << "-->tail" << endl;

	return 0;

}