给定两个链表，创建并集和交集列表，其中包含给定列表中存在的元素的并集和交集。输出列表中元素的顺序无关紧要。
例子：
Input:
List1: 10->15->4->20
lsit2: 8->4->2->10
Output:
Intersection List: 4->10
Union List: 2->8->20->4->15->10方法一（简单）
以下是分别获取并集和交集列表的简单算法。
交叉点（列表1，列表2）
将结果列表初始化为 NULL。遍历 list1 并在 list2 中查找其每个元素，如果该元素存在于 list2 中，则将该元素添加到结果中。
联合（列表1，列表2）：
将结果列表初始化为 NULL。遍历 list1 并将其所有元素添加到结果中。
遍历列表2。如果结果中已经存在 list2 的元素，则不要将其插入到结果中，否则插入。
此方法假定给定列表中没有重复项。
感谢 Shekhu 提出这种方法。以下是此方法的 C 和Java实现。
C++
// C++ program to find union
// and intersection of two unsorted
// linked lists
#include
using namespace std;
/* Link list node */
struct Node {
int data;
struct Node* next;
};
/* A utility function to insert a
node at the beginning ofa linked list*/
void push(struct Node** head_ref, int new_data);
/* A utility function to check if
given data is present in a list */
bool isPresent(struct Node* head, int data);
/* Function to get union of two
linked lists head1 and head2 */
struct Node* getUnion(
struct Node* head1,
struct Node* head2)
{
struct Node* result = NULL;
struct Node *t1 = head1, *t2 = head2;
// Insert all elements of
// list1 to the result list
while (t1 != NULL) {
push(&result, t1->data);
t1 = t1->next;
}
// Insert those elements of list2
// which are not present in result list
while (t2 != NULL) {
if (!isPresent(result, t2->data))
push(&result, t2->data);
t2 = t2->next;
}
return result;
}
/* Function to get intersection of
two linked lists head1 and head2 */
struct Node* getIntersection(struct Node* head1,
struct Node* head2)
{
struct Node* result = NULL;
struct Node* t1 = head1;
// Traverse list1 and search each element of it in
// list2. If the element is present in list 2, then
// insert the element to result
while (t1 != NULL) {
if (isPresent(head2, t1->data))
push(&result, t1->data);
t1 = t1->next;
}
return result;
}
/* A utility function to insert a
node at the beginning of a linked list*/
void push(struct Node** head_ref, int new_data)
{
/* allocate node */
struct Node* new_node
= (struct Node*)malloc(
sizeof(struct Node));
/* put in the data */
new_node->data = new_data;
/* link the old list off the new node */
new_node->next = (*head_ref);
/* move the head to point to the new node */
(*head_ref) = new_node;
}
/* A utility function to print a linked list*/
void printList(struct Node* node)
{
while (node != NULL) {
cout << " " << node->data;
node = node->next;
}
}
/* A utility function that returns true if data is
present in linked list else return false */
bool isPresent(struct Node* head, int data)
{
struct Node* t = head;
while (t != NULL) {
if (t->data == data)
return 1;
t = t->next;
}
return 0;
}
/* Driver program to test above function*/
int main()
{
/* Start with the empty list */
struct Node* head1 = NULL;
struct Node* head2 = NULL;
struct Node* intersecn = NULL;
struct Node* unin = NULL;
/*create a linked lits 10->15->5->20 */
push(&head1, 20);
push(&head1, 4);
push(&head1, 15);
push(&head1, 10);
/*create a linked lits 8->4->2->10 */
push(&head2, 10);
push(&head2, 2);
push(&head2, 4);
push(&head2, 8);
intersecn = getIntersection(head1, head2);
unin = getUnion(head1, head2);
cout << "\n First list is " << endl;
printList(head1);
cout << "\n Second list is " << endl;
printList(head2);
cout << "\n Intersection list is " << endl;
printList(intersecn);
cout << "\n Union list is " << endl;
printList(unin);
return 0;
}
// This code is contributed by shivanisingh2110 C
// C program to find union
// and intersection of two unsorted
// linked lists
#include
#include
#include
/* Link list node */
struct Node {
int data;
struct Node* next;
};
/* A utility function to insert a
node at the beginning ofa linked list*/
void push(struct Node** head_ref, int new_data);
/* A utility function to check if
given data is present in a list */
bool isPresent(struct Node* head, int data);
/* Function to get union of two
linked lists head1 and head2 */
struct Node* getUnion(
struct Node* head1,
struct Node* head2)
{
struct Node* result = NULL;
struct Node *t1 = head1, *t2 = head2;
// Insert all elements of
// list1 to the result list
while (t1 != NULL) {
push(&result, t1->data);
t1 = t1->next;
}
// Insert those elements of list2
// which are not present in result list
while (t2 != NULL) {
if (!isPresent(result, t2->data))
push(&result, t2->data);
t2 = t2->next;
}
return result;
}
/* Function to get intersection of
two linked lists head1 and head2 */
struct Node* getIntersection(struct Node* head1,
struct Node* head2)
{
struct Node* result = NULL;
struct Node* t1 = head1;
// Traverse list1 and search each element of it in
// list2. If the element is present in list 2, then
// insert the element to result
while (t1 != NULL) {
if (isPresent(head2, t1->data))
push(&result, t1->data);
t1 = t1->next;
}
return result;
}
/* A utility function to insert a
node at the beginning of a linked list*/
void push(struct Node** head_ref, int new_data)
{
/* allocate node */
struct Node* new_node
= (struct Node*)malloc(
sizeof(struct Node));
/* put in the data */
new_node->data = new_data;
/* link the old list off the new node */
new_node->next = (*head_ref);
/* move the head to point to the new node */
(*head_ref) = new_node;
}
/* A utility function to print a linked list*/
void printList(struct Node* node)
{
while (node != NULL) {
printf("%d ", node->data);
node = node->next;
}
}
/* A utility function that returns true if data is
present in linked list else return false */
bool isPresent(struct Node* head, int data)
{
struct Node* t = head;
while (t != NULL) {
if (t->data == data)
return 1;
t = t->next;
}
return 0;
}
/* Driver program to test above function*/
int main()
{
/* Start with the empty list */
struct Node* head1 = NULL;
struct Node* head2 = NULL;
struct Node* intersecn = NULL;
struct Node* unin = NULL;
/*create a linked lits 10->15->5->20 */
push(&head1, 20);
push(&head1, 4);
push(&head1, 15);
push(&head1, 10);
/*create a linked lits 8->4->2->10 */
push(&head2, 10);
push(&head2, 2);
push(&head2, 4);
push(&head2, 8);
intersecn = getIntersection(head1, head2);
unin = getUnion(head1, head2);
printf("\n First list is \n");
printList(head1);
printf("\n Second list is \n");
printList(head2);
printf("\n Intersection list is \n");
printList(intersecn);
printf("\n Union list is \n");
printList(unin);
return 0;
} Java
// Java program to find union and
// intersection of two unsorted
// linked lists
class LinkedList {
Node head; // head of list
/* Linked list Node*/
class Node {
int data;
Node next;
Node(int d)
{
data = d;
next = null;
}
}
/* Function to get Union of 2 Linked Lists */
void getUnion(Node head1, Node head2)
{
Node t1 = head1, t2 = head2;
// insert all elements of list1 in the result
while (t1 != null) {
push(t1.data);
t1 = t1.next;
}
// insert those elements of list2
// that are not present
while (t2 != null) {
if (!isPresent(head, t2.data))
push(t2.data);
t2 = t2.next;
}
}
void getIntersection(Node head1, Node head2)
{
Node result = null;
Node t1 = head1;
// Traverse list1 and search each
// element of it in list2.
// If the element is present in
// list 2, then insert the
// element to result
while (t1 != null) {
if (isPresent(head2, t1.data))
push(t1.data);
t1 = t1.next;
}
}
/* Utility function to print list */
void printList()
{
Node temp = head;
while (temp != null) {
System.out.print(temp.data + " ");
temp = temp.next;
}
System.out.println();
}
/* Inserts a node at start of linked list */
void push(int new_data)
{
/* 1 & 2: Allocate the Node &
Put in the data*/
Node new_node = new Node(new_data);
/* 3. Make next of new Node as head */
new_node.next = head;
/* 4. Move the head to point to new Node */
head = new_node;
}
/* A utility function that returns true
if data is present in linked list
else return false */
boolean isPresent(Node head, int data)
{
Node t = head;
while (t != null) {
if (t.data == data)
return true;
t = t.next;
}
return false;
}
/* Driver program to test above functions */
public static void main(String args[])
{
LinkedList llist1 = new LinkedList();
LinkedList llist2 = new LinkedList();
LinkedList unin = new LinkedList();
LinkedList intersecn = new LinkedList();
/*create a linked lits 10->15->5->20 */
llist1.push(20);
llist1.push(4);
llist1.push(15);
llist1.push(10);
/*create a linked lits 8->4->2->10 */
llist2.push(10);
llist2.push(2);
llist2.push(4);
llist2.push(8);
intersecn.getIntersection(llist1.head, llist2.head);
unin.getUnion(llist1.head, llist2.head);
System.out.println("First List is");
llist1.printList();
System.out.println("Second List is");
llist2.printList();
System.out.println("Intersection List is");
intersecn.printList();
System.out.println("Union List is");
unin.printList();
}
} /* This code is contributed by Rajat Mishra */C#
// C# program to find union and
// intersection of two unsorted
// linked lists
using System;
class LinkedList {
public Node head; // head of list
/* Linked list Node*/
public class Node {
public int data;
public Node next;
public Node(int d)
{
data = d;
next = null;
}
}
/* Function to get Union of 2 Linked Lists */
void getUnion(Node head1, Node head2)
{
Node t1 = head1, t2 = head2;
// insert all elements of list1 in the result
while (t1 != null) {
push(t1.data);
t1 = t1.next;
}
// insert those elements of list2
// that are not present
while (t2 != null) {
if (!isPresent(head, t2.data))
push(t2.data);
t2 = t2.next;
}
}
void getIntersection(Node head1, Node head2)
{
Node t1 = head1;
// Traverse list1 and search each
// element of it in list2.
// If the element is present in
// list 2, then insert the
// element to result
while (t1 != null) {
if (isPresent(head2, t1.data))
push(t1.data);
t1 = t1.next;
}
}
/* Utility function to print list */
void printList()
{
Node temp = head;
while (temp != null) {
Console.Write(temp.data + " ");
temp = temp.next;
}
Console.WriteLine();
}
/* Inserts a node at start of linked list */
void push(int new_data)
{
/* 1 & 2: Allocate the Node &
Put in the data*/
Node new_node = new Node(new_data);
/* 3. Make next of new Node as head */
new_node.next = head;
/* 4. Move the head to point to new Node */
head = new_node;
}
/* A utility function that returns true
if data is present in linked list
else return false */
bool isPresent(Node head, int data)
{
Node t = head;
while (t != null) {
if (t.data == data)
return true;
t = t.next;
}
return false;
}
/* Driver code*/
public static void Main(string []args)
{
LinkedList llist1 = new LinkedList();
LinkedList llist2 = new LinkedList();
LinkedList unin = new LinkedList();
LinkedList intersecn = new LinkedList();
/*create a linked lits 10->15->5->20 */
llist1.push(20);
llist1.push(4);
llist1.push(15);
llist1.push(10);
/*create a linked lits 8->4->2->10 */
llist2.push(10);
llist2.push(2);
llist2.push(4);
llist2.push(8);
intersecn.getIntersection(llist1.head, llist2.head);
unin.getUnion(llist1.head, llist2.head);
Console.WriteLine("First List is");
llist1.printList();
Console.WriteLine("Second List is");
llist2.printList();
Console.WriteLine("Intersection List is");
intersecn.printList();
Console.WriteLine("Union List is");
unin.printList();
}
}
// This code is contributed by rutvik_56.Java
// Java code for Union and Intersection of two
// Linked Lists
import java.util.HashMap;
import java.util.HashSet;
class LinkedList {
Node head; // head of list
/* Linked list Node*/
class Node {
int data;
Node next;
Node(int d)
{
data = d;
next = null;
}
}
/* Utility function to print list */
void printList()
{
Node temp = head;
while (temp != null) {
System.out.print(temp.data + " ");
temp = temp.next;
}
System.out.println();
}
/* Inserts a node at start of linked list */
void push(int new_data)
{
/* 1 & 2: Allocate the Node &
Put in the data*/
Node new_node = new Node(new_data);
/* 3. Make next of new Node as head */
new_node.next = head;
/* 4. Move the head to point to new Node */
head = new_node;
}
public void append(int new_data)
{
if (this.head == null) {
Node n = new Node(new_data);
this.head = n;
return;
}
Node n1 = this.head;
Node n2 = new Node(new_data);
while (n1.next != null) {
n1 = n1.next;
}
n1.next = n2;
n2.next = null;
}
/* A utility function that returns true if data is
present in linked list else return false */
boolean isPresent(Node head, int data)
{
Node t = head;
while (t != null) {
if (t.data == data)
return true;
t = t.next;
}
return false;
}
LinkedList getIntersection(Node head1, Node head2)
{
HashSet hset = new HashSet<>();
Node n1 = head1;
Node n2 = head2;
LinkedList result = new LinkedList();
// loop stores all the elements of list1 in hset
while (n1 != null) {
if (hset.contains(n1.data)) {
hset.add(n1.data);
}
else {
hset.add(n1.data);
}
n1 = n1.next;
}
// For every element of list2 present in hset
// loop inserts the element into the result
while (n2 != null) {
if (hset.contains(n2.data)) {
result.push(n2.data);
}
n2 = n2.next;
}
return result;
}
LinkedList getUnion(Node head1, Node head2)
{
// HashMap that will store the
// elements of the lists with their counts
HashMap hmap = new HashMap<>();
Node n1 = head1;
Node n2 = head2;
LinkedList result = new LinkedList();
// loop inserts the elements and the count of
// that element of list1 into the hmap
while (n1 != null) {
if (hmap.containsKey(n1.data)) {
int val = hmap.get(n1.data);
hmap.put(n1.data, val + 1);
}
else {
hmap.put(n1.data, 1);
}
n1 = n1.next;
}
// loop further adds the elements of list2 with
// their counts into the hmap
while (n2 != null) {
if (hmap.containsKey(n2.data)) {
int val = hmap.get(n2.data);
hmap.put(n2.data, val + 1);
}
else {
hmap.put(n2.data, 1);
}
n2 = n2.next;
}
// Eventually add all the elements
// into the result that are present in the hmap
for (int a : hmap.keySet()) {
result.append(a);
}
return result;
}
/* Driver program to test above functions */
public static void main(String args[])
{
LinkedList llist1 = new LinkedList();
LinkedList llist2 = new LinkedList();
LinkedList union = new LinkedList();
LinkedList intersection = new LinkedList();
/*create a linked list 10->15->4->20 */
llist1.push(20);
llist1.push(4);
llist1.push(15);
llist1.push(10);
/*create a linked list 8->4->2->10 */
llist2.push(10);
llist2.push(2);
llist2.push(4);
llist2.push(8);
intersection
= intersection.getIntersection(llist1.head,
llist2.head);
union = union.getUnion(llist1.head, llist2.head);
System.out.println("First List is");
llist1.printList();
System.out.println("Second List is");
llist2.printList();
System.out.println("Intersection List is");
intersection.printList();
System.out.println("Union List is");
union.printList();
}
}
// This code is contributed by Kamal Rawal C#
// C# code for Union and Intersection of two
// Linked Lists
using System;
using System.Collections;
using System.Collections.Generic;
class LinkedList
{
public Node head; // head of list
/* Linked list Node*/
public class Node
{
public int data;
public Node next;
public Node(int d)
{
data = d;
next = null;
}
}
/* Utility function to print list */
void printList()
{
Node temp = head;
while (temp != null)
{
Console.Write(temp.data + " ");
temp = temp.next;
}
Console.WriteLine();
}
/* Inserts a node at start of linked list */
void push(int new_data)
{
/* 1 & 2: Allocate the Node &
Put in the data*/
Node new_node = new Node(new_data);
/* 3. Make next of new Node as head */
new_node.next = head;
/* 4. Move the head to point to new Node */
head = new_node;
}
public void append(int new_data)
{
if (this.head == null) {
Node n = new Node(new_data);
this.head = n;
return;
}
Node n1 = this.head;
Node n2 = new Node(new_data);
while (n1.next != null) {
n1 = n1.next;
}
n1.next = n2;
n2.next = null;
}
/* A utility function that returns true if data is
present in linked list else return false */
bool isPresent(Node head, int data)
{
Node t = head;
while (t != null) {
if (t.data == data)
return true;
t = t.next;
}
return false;
}
LinkedList getIntersection(Node head1, Node head2)
{
HashSet hset = new HashSet();
Node n1 = head1;
Node n2 = head2;
LinkedList result = new LinkedList();
// loop stores all the elements of list1 in hset
while (n1 != null) {
if (hset.Contains(n1.data)) {
hset.Add(n1.data);
}
else {
hset.Add(n1.data);
}
n1 = n1.next;
}
// For every element of list2 present in hset
// loop inserts the element into the result
while (n2 != null) {
if (hset.Contains(n2.data)) {
result.push(n2.data);
}
n2 = n2.next;
}
return result;
}
LinkedList getUnion(Node head1, Node head2)
{
// HashMap that will store the
// elements of the lists with their counts
SortedDictionary hmap = new SortedDictionary();
Node n1 = head1;
Node n2 = head2;
LinkedList result = new LinkedList();
// loop inserts the elements and the count of
// that element of list1 into the hmap
while (n1 != null) {
if (hmap.ContainsKey(n1.data)) {
hmap[n1.data]++;
}
else {
hmap[n1.data]= 1;
}
n1 = n1.next;
}
// loop further adds the elements of list2 with
// their counts into the hmap
while (n2 != null) {
if (hmap.ContainsKey(n2.data)) {
hmap[n2.data]++;
}
else {
hmap[n2.data]= 1;
}
n2 = n2.next;
}
// Eventually add all the elements
// into the result that are present in the hmap
foreach(int a in hmap.Keys) {
result.append(a);
}
return result;
}
/* Driver program to test above functions */
public static void Main(string []args)
{
LinkedList llist1 = new LinkedList();
LinkedList llist2 = new LinkedList();
LinkedList union = new LinkedList();
LinkedList intersection = new LinkedList();
/*create a linked list 10->15->4->20 */
llist1.push(20);
llist1.push(4);
llist1.push(15);
llist1.push(10);
/*create a linked list 8->4->2->10 */
llist2.push(10);
llist2.push(2);
llist2.push(4);
llist2.push(8);
intersection = intersection.getIntersection(llist1.head,
llist2.head);
union = union.getUnion(llist1.head, llist2.head);
Console.WriteLine("First List is");
llist1.printList();
Console.WriteLine("Second List is");
llist2.printList();
Console.WriteLine("Intersection List is");
intersection.printList();
Console.WriteLine("Union List is");
union.printList();
}
}
//This code is contributed by pratham76 输出
First list is
10 15 4 20
Second list is
8 4 2 10
Intersection list is
4 10
Union list is
2 8 20 4 15 10复杂度分析：
- 时间复杂度： O(m*n)。
这里 ‘m’ 和 ‘n’ 分别是第一个和第二个列表中存在的元素数。
对于联合：对于 list-2 中的每个元素，我们检查该元素是否已经存在于使用 list-1 生成的结果列表中。
对于交集：对于 list-1 中的每个元素，我们检查该元素是否也存在于 list-2 中。 - 辅助空间： O(1)。
不使用任何数据结构来存储值。
方法二（使用归并排序）
在这种方法中，Union 和 Intersection 的算法非常相似。首先，我们对给定的列表进行排序，然后遍历已排序的列表以获取并集和交集。
以下是获取并集和交集列表要遵循的步骤。
- 使用归并排序对第一个链表进行排序。这一步需要 O(mLogm) 时间。有关此步骤的详细信息，请参阅此帖子。
- 使用归并排序对第二个链表进行排序。这一步需要 O(nLogn) 时间。有关此步骤的详细信息，请参阅此帖子。
- 线性扫描两个排序列表以获得并集和交集。这一步需要 O(m + n) 时间。此步骤可以使用与此处讨论的排序数组算法相同的算法来实现。
该方法的时间复杂度为 O(mLogm + nLogn)，优于方法 1 的时间复杂度。
方法三（使用哈希）
联合 (list1, list2)
将结果列表初始化为 NULL 并创建一个空的哈希表。一一遍历两个列表，对于每个被访问的元素，在哈希表中查找该元素。如果该元素不存在，则将该元素插入到结果列表中。如果该元素存在，则忽略它。
交叉点（列表1，列表2）
将结果列表初始化为 NULL 并创建一个空的哈希表。遍历列表1。对于 list1 中被访问的每个元素，将元素插入哈希表中。遍历list2，对于list2中被访问的每个元素，在哈希表中查找该元素。如果该元素存在，则将该元素插入到结果列表中。如果该元素不存在，则忽略它。
以上两种方法都假设没有重复项。
Java
// Java code for Union and Intersection of two
// Linked Lists
import java.util.HashMap;
import java.util.HashSet;
class LinkedList {
Node head; // head of list
/* Linked list Node*/
class Node {
int data;
Node next;
Node(int d)
{
data = d;
next = null;
}
}
/* Utility function to print list */
void printList()
{
Node temp = head;
while (temp != null) {
System.out.print(temp.data + " ");
temp = temp.next;
}
System.out.println();
}
/* Inserts a node at start of linked list */
void push(int new_data)
{
/* 1 & 2: Allocate the Node &
Put in the data*/
Node new_node = new Node(new_data);
/* 3. Make next of new Node as head */
new_node.next = head;
/* 4. Move the head to point to new Node */
head = new_node;
}
public void append(int new_data)
{
if (this.head == null) {
Node n = new Node(new_data);
this.head = n;
return;
}
Node n1 = this.head;
Node n2 = new Node(new_data);
while (n1.next != null) {
n1 = n1.next;
}
n1.next = n2;
n2.next = null;
}
/* A utility function that returns true if data is
present in linked list else return false */
boolean isPresent(Node head, int data)
{
Node t = head;
while (t != null) {
if (t.data == data)
return true;
t = t.next;
}
return false;
}
LinkedList getIntersection(Node head1, Node head2)
{
HashSet hset = new HashSet<>();
Node n1 = head1;
Node n2 = head2;
LinkedList result = new LinkedList();
// loop stores all the elements of list1 in hset
while (n1 != null) {
if (hset.contains(n1.data)) {
hset.add(n1.data);
}
else {
hset.add(n1.data);
}
n1 = n1.next;
}
// For every element of list2 present in hset
// loop inserts the element into the result
while (n2 != null) {
if (hset.contains(n2.data)) {
result.push(n2.data);
}
n2 = n2.next;
}
return result;
}
LinkedList getUnion(Node head1, Node head2)
{
// HashMap that will store the
// elements of the lists with their counts
HashMap hmap = new HashMap<>();
Node n1 = head1;
Node n2 = head2;
LinkedList result = new LinkedList();
// loop inserts the elements and the count of
// that element of list1 into the hmap
while (n1 != null) {
if (hmap.containsKey(n1.data)) {
int val = hmap.get(n1.data);
hmap.put(n1.data, val + 1);
}
else {
hmap.put(n1.data, 1);
}
n1 = n1.next;
}
// loop further adds the elements of list2 with
// their counts into the hmap
while (n2 != null) {
if (hmap.containsKey(n2.data)) {
int val = hmap.get(n2.data);
hmap.put(n2.data, val + 1);
}
else {
hmap.put(n2.data, 1);
}
n2 = n2.next;
}
// Eventually add all the elements
// into the result that are present in the hmap
for (int a : hmap.keySet()) {
result.append(a);
}
return result;
}
/* Driver program to test above functions */
public static void main(String args[])
{
LinkedList llist1 = new LinkedList();
LinkedList llist2 = new LinkedList();
LinkedList union = new LinkedList();
LinkedList intersection = new LinkedList();
/*create a linked list 10->15->4->20 */
llist1.push(20);
llist1.push(4);
llist1.push(15);
llist1.push(10);
/*create a linked list 8->4->2->10 */
llist2.push(10);
llist2.push(2);
llist2.push(4);
llist2.push(8);
intersection
= intersection.getIntersection(llist1.head,
llist2.head);
union = union.getUnion(llist1.head, llist2.head);
System.out.println("First List is");
llist1.printList();
System.out.println("Second List is");
llist2.printList();
System.out.println("Intersection List is");
intersection.printList();
System.out.println("Union List is");
union.printList();
}
}
// This code is contributed by Kamal Rawal
C＃
// C# code for Union and Intersection of two
// Linked Lists
using System;
using System.Collections;
using System.Collections.Generic;
class LinkedList
{
public Node head; // head of list
/* Linked list Node*/
public class Node
{
public int data;
public Node next;
public Node(int d)
{
data = d;
next = null;
}
}
/* Utility function to print list */
void printList()
{
Node temp = head;
while (temp != null)
{
Console.Write(temp.data + " ");
temp = temp.next;
}
Console.WriteLine();
}
/* Inserts a node at start of linked list */
void push(int new_data)
{
/* 1 & 2: Allocate the Node &
Put in the data*/
Node new_node = new Node(new_data);
/* 3. Make next of new Node as head */
new_node.next = head;
/* 4. Move the head to point to new Node */
head = new_node;
}
public void append(int new_data)
{
if (this.head == null) {
Node n = new Node(new_data);
this.head = n;
return;
}
Node n1 = this.head;
Node n2 = new Node(new_data);
while (n1.next != null) {
n1 = n1.next;
}
n1.next = n2;
n2.next = null;
}
/* A utility function that returns true if data is
present in linked list else return false */
bool isPresent(Node head, int data)
{
Node t = head;
while (t != null) {
if (t.data == data)
return true;
t = t.next;
}
return false;
}
LinkedList getIntersection(Node head1, Node head2)
{
HashSet hset = new HashSet();
Node n1 = head1;
Node n2 = head2;
LinkedList result = new LinkedList();
// loop stores all the elements of list1 in hset
while (n1 != null) {
if (hset.Contains(n1.data)) {
hset.Add(n1.data);
}
else {
hset.Add(n1.data);
}
n1 = n1.next;
}
// For every element of list2 present in hset
// loop inserts the element into the result
while (n2 != null) {
if (hset.Contains(n2.data)) {
result.push(n2.data);
}
n2 = n2.next;
}
return result;
}
LinkedList getUnion(Node head1, Node head2)
{
// HashMap that will store the
// elements of the lists with their counts
SortedDictionary hmap = new SortedDictionary();
Node n1 = head1;
Node n2 = head2;
LinkedList result = new LinkedList();
// loop inserts the elements and the count of
// that element of list1 into the hmap
while (n1 != null) {
if (hmap.ContainsKey(n1.data)) {
hmap[n1.data]++;
}
else {
hmap[n1.data]= 1;
}
n1 = n1.next;
}
// loop further adds the elements of list2 with
// their counts into the hmap
while (n2 != null) {
if (hmap.ContainsKey(n2.data)) {
hmap[n2.data]++;
}
else {
hmap[n2.data]= 1;
}
n2 = n2.next;
}
// Eventually add all the elements
// into the result that are present in the hmap
foreach(int a in hmap.Keys) {
result.append(a);
}
return result;
}
/* Driver program to test above functions */
public static void Main(string []args)
{
LinkedList llist1 = new LinkedList();
LinkedList llist2 = new LinkedList();
LinkedList union = new LinkedList();
LinkedList intersection = new LinkedList();
/*create a linked list 10->15->4->20 */
llist1.push(20);
llist1.push(4);
llist1.push(15);
llist1.push(10);
/*create a linked list 8->4->2->10 */
llist2.push(10);
llist2.push(2);
llist2.push(4);
llist2.push(8);
intersection = intersection.getIntersection(llist1.head,
llist2.head);
union = union.getUnion(llist1.head, llist2.head);
Console.WriteLine("First List is");
llist1.printList();
Console.WriteLine("Second List is");
llist2.printList();
Console.WriteLine("Intersection List is");
intersection.printList();
Console.WriteLine("Union List is");
union.printList();
}
}
//This code is contributed by pratham76
输出
First List is
10 15 4 20
Second List is
8 4 2 10
Intersection List is
10 4
Union List is
2 4 20 8 10 15复杂度分析：
- 时间复杂度： O(m+n)。
这里 ‘m’ 和 ‘n’ 分别是第一个和第二个列表中存在的元素数。
对于联合：遍历两个列表，将元素存储在 Hash-map 中并更新各自的计数。
对于交集：首先遍历 list-1，将其元素存储在 Hash-map 中，然后对 list-2 中的每个元素检查它是否已经存在于映射中。这需要 O(1) 时间。 - 辅助空间： O(m+n)。
使用 Hash-map 数据结构来存储值。
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