两个链表求交集_实现两个排序链表的并集和交集

两个链表求交集

In computer science, a linked list is a linear collection of data elements, whose order is not given by their physical placement in memory. Instead, each element points to the next. It is a data structure consisting of a collection of nodes which together represent a sequence. In its most basic form, each node contains data and a reference (in other words, a link) to the next node in the sequence. This structure allows for efficient insertion or removal of elements from any position in the sequence during iteration. More complex variants add additional links, allowing more efficient insertion or removal of nodes at arbitrary positions.

在计算机科学中，链表是数据元素的线性集合，其顺序不是由它们在内存中的物理位置给出的。 相反，每个元素都指向下一个。 它是一种数据结构，由节点的集合组成，这些节点一起代表一个序列。 以其最基本的形式，每个节点都包含数据和对序列中下一个节点的引用(换句话说，就是链接)。 这种结构允许在迭代过程中从序列中的任何位置有效插入或删除元素。 更复杂的变体会添加其他链接，从而可以更有效地在任意位置插入或删除节点。

Union of two linked lists can be found by using merging the lists in a sorted manner.

两个链接列表的联盟可以通过合并列表的排序方式找到。

The intersection of the two lists can be found by only taking common elements while merging the two lists.

通过合并两个列表时仅采用公共元素，可以找到两个列表的交集 。

It has been assumed that the linked lists are sorted.

假定已对链表进行排序 。

Algorithm:

算法：

Union(L1,L2)

联合(L1，L2)

    1) Declare node pointer output, output Tail as NULL
    2) Repeat steps 3 to 9 while L1!=NULL AND L2!=NULL
    3) Make a newNode and set its next = NULL
    4) If L1->data < L2->data then
        Set newNode->data = L1->data
        Set L1 = L1->next
    5) Else if L1->data > L2->data then
        Set newNode->data = L2->data
        L2 = L2->next
    6) Else
        i) Set Data = L1->data
        ii) Set newNode->data = Data
        iii) Repeat steps a) and b) 
        while L1!=NULL AND L2!=NULL AND L1->data == Data AND L2->data == Data
            a) Set L1 = L1->next
            b) Set L2 = L2->next

    7) If output == NULL then
        Set Output = outputTail = newNode

    8) Else
        a) Set outputTail->next = newNode
        b) Set outputTail = outputTail->next
     9) Repeat steps 10 to 14 while L1!=NULL
    10) Make a newNode
    11) Set outputTail->next = newNode
    12) Set outputTail = outputTail->next
    13) Set outputTail->data = L1->data
    14) Set L1 = L1->next
        Repeat steps 15 to 19 while L2!=NULL
    15) Make a newNode
    16) Set outputTail->next = newNode
    17) Set outputTail = outputTail->next
    18) Set outputTail->data = L2->data
    19) Set L2 = L2->next
        Return output

Intersection(L1,L2)

交叉路口(L1，L2)

    1.	If L1 or L2 is NULL then return NULL
    2.	Declare node pointers output, outputTail as null.
    3.	Repeat steps 4 to 6 while L1!=NULL AND L2!=NULL
    4.	If L1->data
   
    data then
        Set L1 = L1->next
    5.	Else If L2->data
    
     data then
        Set L2 = L2->next
    6.	Else
        a)	Declare and set data = L1->data
        b)	Make a newNode
        c)	Set newNode->data = data and newNode->next = NULL
        d)	If output == null then
            i.	Set output = outputTail = newNode
        e)	Else
            i.	Set outputTail->next = newNode
            ii.	Set outputTail = outputTail->next
        f)	Repeat steps i and ii 
        while L1!=NULL AND L2!=NULL AND L1->data == data AND L2->data == data
            i.	Set L1 = L1->next
            ii.	Set L2 = L2->next

    
   

Code:

码：

#include <stdio.h>
#include <stdlib.h>

struct node{
	struct node*next;
	int data;
};

struct node * Union(struct node * L1, struct node * L2){
	struct node * output = NULL;
	struct node * outTail = NULL;
	while(L1&&L2){
		struct node * newNode = (struct node *) malloc(sizeof(struct node));
		newNode->next = NULL;
		if(L1->data<L2->data){
			newNode->data = L1->data;
			L1 = L1->next;
		}
		else if(L1->data>L2->data){
			newNode->data = L2->data;
			L2 = L2->next;
		}
		else{
			int data = L1->data;
			newNode->data = data;
			while(L1 && L2 && L1->data == data && L2->data == data){
				L1 = L1->next;
				L2 = L2->next;
			}
		}
		if(!output)
			output = outTail = newNode;
		else{
			outTail->next = newNode;
			outTail = outTail->next;
		}
	}
	while(L1){
		outTail->next = (struct node *) malloc(sizeof(struct node));
		outTail = outTail->next;
		outTail->data = L1->data;
		L1 = L1->next;
	}
	while(L2){
		outTail->next = (struct node *) malloc(sizeof(struct node));
		outTail = outTail->next;
		outTail->data = L2->data;
		L2 = L2->next;
	}
	outTail->next = NULL;
	return output;
}

struct node * intersection(struct node * L1, struct node* L2){
	if(L1 == NULL || L2 == NULL)
		return NULL;
	struct node * output = NULL;
	struct node * outTail = NULL;
	while(L1&&L2){
		if(L1->data<L2->data){
			L1 = L1->next;
		}
		else if(L2->data<L1->data){
			L2 = L2->next;
		}
		else{
			int data = L1->data;
			struct node * newNode = (struct node *) malloc(sizeof(struct node));
			newNode->data = data;
			newNode->next = NULL;
			if(output == NULL){
				outTail = output = newNode;
			}
			else{
				outTail->next = newNode;
				outTail = outTail->next;
			}
			while(L1 && L2 && L1->data == data && L2->data == data){
				L1 = L1->next;
				L2 = L2->next;
			}
		}
	}
	return output;
}

struct node * createList(int listNum){
	struct node * list = NULL;
	struct node * list_tail = NULL;
	printf("Enter elements of List %d in increasing order\n",listNum);
	char ch = 'y';
	do{
		int data;
		printf("Enter element : ");
		scanf("%d",&data);
		struct node * newNode = (struct node *) malloc(sizeof(struct node));
		newNode->data = data;
		newNode->next = NULL;
		if(list == NULL){
			list = list_tail = newNode;
		}
		else{
			list_tail->next = newNode;
			list_tail = list_tail->next;
		}
		printf("Would you like to insert another element [Y/N] : ");
		scanf(" %c",&ch);
	}while(ch == 'y' || ch == 'Y');
	
	return list;
}

void print(struct node * list){
	if(list == NULL){
		printf("Empty List\n");
		return;
	}
	while(list!=NULL){
		printf("%d ",list->data);
		list = list->next;
	}
	printf("\n");
}

int main() {
	struct node * L1 = NULL;
	struct node * L2 = NULL;
	struct node * L3 = NULL;
	struct node * L4 = NULL;
	L1 = createList(1);
	L2 = createList(2);
	printf("List 1 : ");
	print(L1);
	printf("List 2 : ");
	print(L2);
	printf("Union : ");
	L3 = Union(L1, L2);
	print(L3);
	printf("Intersection : ");
	L4 = intersection(L1, L2);
	print(L4);
	return 0;
}

Output

输出量

Enter elements of List 1 in increasing order 
Enter element : 1
Would you like to insert another element [Y/N] : Y 
Enter element : 4
Would you like to insert another element [Y/N] : Y 
Enter element : 9
Would you like to insert another element [Y/N] : Y 
Enter element : 27 
Would you like to insert another element [Y/N] : N 
Enter elements of List 2 in increasing order 
Enter element : 4
Would you like to insert another element [Y/N] : Y 
Enter element : 9
Would you like to insert another element [Y/N] : Y 
Enter element : 18 
Would you like to insert another element [Y/N] : Y 
Enter element : 22 
Would you like to insert another element [Y/N] : Y 
Enter element : 30 
Would you like to insert another element [Y/N] : N 
List 1 : 1 4 9 27
List 2 : 4 9 18 22 30
Union : 1 4 9 18 22 27 30
Intersection : 4 9 

翻译自: https://www.includehelp.com/data-structure-tutorial/implement-union-and-intersection-of-two-sorted-linked-lists.aspx

两个链表求交集