从两个链表中计算总和等于给定值的对

给定两个不同元素的大小为n1和n2的链表（可以排序或未排序）。给定一个值x 。问题是计算两个列表中总和等于给定值x 的所有对。
注意：该对从每个链表中都有一个元素。
例子：

Input : list1 = 3->1->5->7
        list2 = 8->2->5->3
        x = 10
Output : 2
The pairs are:
(5, 5) and (7, 3)

Input : list1 = 4->3->5->7->11->2->1
        list2 = 2->3->4->5->6->8-12
        x = 9         
Output : 5

方法 1（朴素方法）：使用两个循环从两个链表中选取元素并检查对的总和是否等于x 。

C/C++

// C++ implementation to count pairs from both linked 
// lists  whose sum is equal to a given value
#include 
using namespace std;

/* A Linked list node */
struct Node
{
  int data;
  struct Node* next;
};

// function to insert a node at the
// beginning of the 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 to the new node */
  new_node->next = (*head_ref);
 
  /* move the head to point to the new node */
  (*head_ref)    = new_node;
}

// function to count all pairs from both the linked lists
// whose sum is equal to a given value
int countPairs(struct Node* head1, struct Node* head2, int x)
{
    int count = 0;
    
    struct Node *p1, *p2;
    
    // traverse the 1st linked list
    for (p1 = head1; p1 != NULL; p1 = p1->next)

        // for each node of 1st list
        // traverse the 2nd list

        for (p2 = head2; p2 != NULL; p2 = p2->next)

            // if sum of pair is equal to 'x'
            // increment count
            if ((p1->data + p2->data) == x)
                count++;            
        
    // required count of pairs     
    return count;
}

// Driver program to test above
int main()
{
    struct Node* head1 = NULL;
    struct Node* head2 = NULL;
    
    // create linked list1 3->1->5->7
    push(&head1, 7);
    push(&head1, 5);
    push(&head1, 1);
    push(&head1, 3);    
    
    // create linked list2 8->2->5->3
    push(&head2, 3);
    push(&head2, 5);
    push(&head2, 2);
    push(&head2, 8);
    
    int x = 10;
    
    cout << "Count = "
         << countPairs(head1, head2, x);
    return 0;
} 

Java

// Java implementation to count pairs from both linked
// lists  whose sum is equal to a given value
 
// Note : here we use java.util.LinkedList for
// linked list implementation
 
import java.util.Arrays;
import java.util.Iterator;
import java.util.LinkedList;
 
class GFG
{
    // method to count all pairs from both the linked lists
    // whose sum is equal to a given value
    static int countPairs(LinkedList head1, LinkedList head2, int x)
    {
        int count = 0;
          
        // traverse the 1st linked list
        Iterator itr1 = head1.iterator();
        while(itr1.hasNext())
        {
            // for each node of 1st list
            // traverse the 2nd list
            Iterator itr2 = head2.iterator();
            Integer t = itr1.next();
            while(itr2.hasNext())
            {
                // if sum of pair is equal to 'x'
                // increment count
                if ((t + itr2.next()) == x)
                    count++;
            }
        }
                            
        // required count of pairs    
        return count;
    }
     
    // Driver method
    public static void main(String[] args)
    {
        Integer arr1[] = {3, 1, 5, 7};
        Integer arr2[] = {8, 2, 5, 3};
         
        // create linked list1 3->1->5->7
        LinkedList head1 = new LinkedList<>(Arrays.asList(arr1));
         
        // create linked list2 8->2->5->3
        LinkedList head2 = new LinkedList<>(Arrays.asList(arr2));
        
        int x = 10;
          
        System.out.println("Count = " + countPairs(head1, head2, x));
    }   
}

Python3

# Python3 implementation to count pairs from both linked
# lists whose sum is equal to a given value
 
# A Linked list node
class Node:
    def __init__(self,data):
        self.data = data
        self.next = None
 
# function to insert a node at the
# beginning of the linked list
 
def push(head_ref,new_data):
    new_node=Node(new_data)
    #new_node.data = new_data
    new_node.next = head_ref
    head_ref = new_node
    return head_ref
 
 
# function to count all pairs from both the linked lists
# whose sum is equal to a given value
def countPairs(head1, head2, x):
    count = 0
     
    #struct Node p1, p2
     
    # traverse the 1st linked list
    p1 = head1
    while(p1 != None):
 
        # for each node of 1st list
        # traverse the 2nd list
        p2 = head2
        while(p2 != None):
 
            # if sum of pair is equal to 'x'
            # increment count
            if ((p1.data + p2.data) == x):
                count+=1
            p2 = p2.next
             
        p1 = p1.next
    # required count of pairs    
    return count
 
 
# Driver program to test above
if __name__=='__main__':
 
    head1 = None
    head2 = None
     
    # create linked list1 3.1.5.7
    head1=push(head1, 7)
    head1=push(head1, 5)
    head1=push(head1, 1)
    head1=push(head1, 3)
     
    # create linked list2 8.2.5.3
    head2=push(head2, 3)
    head2=push(head2, 5)
    head2=push(head2, 2)
    head2=push(head2, 8)
     
    x = 10
     
    print("Count = ",countPairs(head1, head2, x))
     
# This code is contributed by AbhiThakur

C#

// C# implementation to count pairs from both linked
// lists whose sum is equal to a given value
using System;
using System.Collections.Generic;
     
 
// Note : here we use using System.Collections.Generic for
// linked list implementation
class GFG
{
    // method to count all pairs from both the linked lists
    // whose sum is equal to a given value
    static int countPairs(List head1, List head2, int x)
    {
        int count = 0;
         
        // traverse the 1st linked list
         
        foreach(int itr1 in head1)
        {
            // for each node of 1st list
            // traverse the 2nd list
            int t = itr1;
            foreach(int itr2 in head2)
            {
                // if sum of pair is equal to 'x'
                // increment count
                if ((t + itr2) == x)
                    count++;
            }
        }
                             
        // required count of pairs    
        return count;
    }
     
    // Driver code
    public static void Main(String []args)
    {
        int []arr1 = {3, 1, 5, 7};
        int []arr2 = {8, 2, 5, 3};
         
        // create linked list1 3->1->5->7
        List head1 = new List(arr1);
         
        // create linked list2 8->2->5->3
        List head2 = new List(arr2);
         
        int x = 10;
         
    Console.WriteLine("Count = " + countPairs(head1, head2, x));
    }
}
 
// This code is contributed by 29AjayKumar

输出：

Count = 2

时间复杂度： O(n1*n2)
辅助空间： O(1)方法二（排序）：使用归并排序技术将第一个链表升序排序，第二个链表降序排序。现在按以下方式从左到右遍历两个列表：
算法：

countPairs(list1, list2, x)
  Initialize count = 0
  while list != NULL and list2 != NULL
     if (list1->data + list2->data) == x
        list1 = list1->next    
        list2 = list2->next
        count++
    else if (list1->data + list2->data) > x
        list2 = list2->next
    else
        list1 = list1->next

  return count     

为简单起见，下面给出的实现假设 list1 按升序排序， list2 按降序排序。

C/C++

// C++ implementation to count pairs from both linked 
// lists  whose sum is equal to a given value
#include 
using namespace std;

/* A Linked list node */
struct Node
{
  int data;
  struct Node* next;
};

// function to insert a node at the
// beginning of the 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 to the new node */
  new_node->next = (*head_ref);
 
  /* move the head to point to the new node */
  (*head_ref)    = new_node;
}

// function to count all pairs from both the linked 
// lists whose sum is equal to a given value
int countPairs(struct Node* head1, struct Node* head2,
                                              int x)
{
    int count = 0;
    
    // sort head1 in ascending order and
    // head2 in descending order
    // sort (head1), sort (head2)
    // For simplicity both lists are considered to be 
    // sorted in the respective orders
    
    // traverse both the lists from left to right
    while (head1 != NULL && head2 != NULL)
    {
        // if this sum is equal to 'x', then move both 
        // the lists to next nodes and increment 'count'
        if ((head1->data + head2->data) == x)
        {
            head1 = head1->next;
            head2 = head2->next;
            count++;    
        }    
        
        // if this sum is greater than x, then
        // move head2 to next node
        else if ((head1->data + head2->data) > x)
            head2 = head2->next;
            
        // else move head1 to next node    
        else
            head1 = head1->next;
    }        
        
    // required count of pairs     
    return count;
}

// Driver program to test above
int main()
{
    struct Node* head1 = NULL;
    struct Node* head2 = NULL;
    
    // create linked list1 1->3->5->7
    // assumed to be in ascending order
    push(&head1, 7);
    push(&head1, 5);
    push(&head1, 3);
    push(&head1, 1);    
    
    // create linked list2 8->5->3->2
    // assumed to be in descending order
    push(&head2, 2);
    push(&head2, 3);
    push(&head2, 5);
    push(&head2, 8);
    
    int x = 10;
    
    cout << "Count = "
         << countPairs(head1, head2, x);
    return 0;
}

Java

// Java implementation to count pairs from both linked
// lists  whose sum is equal to a given value
 
// Note : here we use java.util.LinkedList for
// linked list implementation
 
import java.util.Arrays;
import java.util.Collections;
import java.util.Iterator;
import java.util.LinkedList;
 
class GFG
{
    // method to count all pairs from both the linked lists
    // whose sum is equal to a given value
    static int countPairs(LinkedList head1, LinkedList head2, int x)
    {
        int count = 0;
          
        // sort head1 in ascending order and
        // head2 in descending order
        Collections.sort(head1);
        Collections.sort(head2,Collections.reverseOrder());
         
        // traverse both the lists from left to right
        Iterator itr1 = head1.iterator();
        Iterator itr2 = head2.iterator();
         
        Integer num1 = itr1.hasNext() ? itr1.next() : null;
        Integer num2 = itr2.hasNext() ? itr2.next() : null;
         
        while(num1 != null && num2 != null)
        {    
             
            // if this sum is equal to 'x', then move both
            // the lists to next nodes and increment 'count'
             
            if ((num1 + num2) == x)
            {
                num1 = itr1.hasNext() ? itr1.next() : null;
                num2 = itr2.hasNext() ? itr2.next() : null;
                 
                count++;
            }
             
            // if this sum is greater than x, then
            // move itr2 to next node
            else if ((num1 + num2) > x)
                num2 = itr2.hasNext() ? itr2.next() : null;
             
            // else move itr1 to next node
            else
                num1 = itr1.hasNext() ? itr1.next() : null;
             
        }
                            
        // required count of pairs    
        return count;
    }
     
    // Driver method
    public static void main(String[] args)
    {
        Integer arr1[] = {3, 1, 5, 7};
        Integer arr2[] = {8, 2, 5, 3};
         
        // create linked list1 3->1->5->7
        LinkedList head1 = new LinkedList<>(Arrays.asList(arr1));
         
        // create linked list2 8->2->5->3
        LinkedList head2 = new LinkedList<>(Arrays.asList(arr2));
        
        int x = 10;
          
        System.out.println("Count = " + countPairs(head1, head2, x));
    }   
}

输出：

Count = 2

时间复杂度： O(n1*logn1) + O(n2*logn2)
辅助空间： O(1)
排序会改变节点的顺序。如果顺序很重要，则可以创建和使用链表的副本。
方法三（Hashing）： C++中使用unordered_set实现哈希表。我们将所有第一个链表元素存储在哈希表中。对于第二个链表的元素，我们从x 中减去每个元素并在哈希表中检查结果。如果结果存在，我们增加计数。

C++

// C++ implementation to count pairs from both linked 
// lists whose sum is equal to a given value
#include 
using namespace std;
 
/* A Linked list node */
struct Node
{
  int data;
  struct Node* next;
};
 
// function to insert a node at the
// beginning of the 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 to the new node */
  new_node->next = (*head_ref);
  
  /* move the head to point to the new node */
  (*head_ref)    = new_node;
}
 
// function to count all pairs from both the linked
// lists whose sum is equal to a given value
int countPairs(struct Node* head1, struct Node* head2,
                                               int x)
{
    int count = 0;
     
    unordered_set us;
     
    // insert all the elements of 1st list
    // in the hash table(unordered_set 'us')
    while (head1 != NULL)
    {
        us.insert(head1->data);   
         
        // move to next node   
        head1 = head1->next;
    }
     
    // for each element of 2nd list
    while (head2 != NULL)   
    {
        // find (x - head2->data) in 'us'
        if (us.find(x - head2->data) != us.end())
            count++;
         
        // move to next node
        head2 = head2->next;   
    }
    // required count of pairs    
    return count;
}
 
// Driver program to test above
int main()
{
    struct Node* head1 = NULL;
    struct Node* head2 = NULL;
     
    // create linked list1 3->1->5->7
    push(&head1, 7);
    push(&head1, 5);
    push(&head1, 1);
    push(&head1, 3);   
     
    // create linked list2 8->2->5->3
    push(&head2, 3);
    push(&head2, 5);
    push(&head2, 2);
    push(&head2, 8);
     
    int x = 10;
     
    cout << "Count = "
         << countPairs(head1, head2, x);
    return 0;
}

Java

// Java implementation to count pairs from both linked
// lists  whose sum is equal to a given value
 
// Note : here we use java.util.LinkedList for
// linked list implementation
 
import java.util.Arrays;
import java.util.HashSet;
import java.util.Iterator;
import java.util.LinkedList;
 
class GFG
{
    // method to count all pairs from both the linked lists
    // whose sum is equal to a given value
    static int countPairs(LinkedList head1, LinkedList head2, int x)
    {
        int count = 0;
          
        HashSet us = new HashSet();
          
        // insert all the elements of 1st list
        // in the hash table(unordered_set 'us')
        Iterator itr1 = head1.iterator();
        while (itr1.hasNext())
        {
            us.add(itr1.next());   
            
        }
         
        Iterator itr2 = head2.iterator();
        // for each element of 2nd list
        while (itr2.hasNext())   
        {
            // find (x - head2->data) in 'us'
            if (!(us.add(x - itr2.next())))
                count++;
                
        }
         
        // required count of pairs    
        return count;
    }
     
    // Driver method
    public static void main(String[] args)
    {
        Integer arr1[] = {3, 1, 5, 7};
        Integer arr2[] = {8, 2, 5, 3};
         
        // create linked list1 3->1->5->7
        LinkedList head1 = new LinkedList<>(Arrays.asList(arr1));
         
        // create linked list2 8->2->5->3
        LinkedList head2 = new LinkedList<>(Arrays.asList(arr2));
        
        int x = 10;
          
        System.out.println("Count = " + countPairs(head1, head2, x));
    }   
}

蟒蛇3

# Python3 implementation to count pairs from both linked 
# lists whose sum is equal to a given value
  
''' A Linked list node '''
class Node:
     
    def __init__(self):
        self.data = 0
        self.next = None
   
# function to add a node at the
# beginning of the linked list
def push(head_ref, new_data):
 
  ''' allocate node '''
  new_node =Node()
   
  ''' put in the data  '''
  new_node.data  = new_data;
   
  ''' link the old list to the new node '''
  new_node.next = (head_ref);
   
  ''' move the head to point to the new node '''
  (head_ref) = new_node;
   
  return head_ref
 
# function to count all pairs from both the linked
# lists whose sum is equal to a given value
def countPairs(head1, head2, x):
    count = 0;   
    us = set()
      
    # add all the elements of 1st list
    # in the hash table(unordered_set 'us')
    while (head1 != None):  
        us.add(head1.data);   
          
        # move to next node   
        head1 = head1.next;
      
    # for each element of 2nd list
    while (head2 != None): 
     
        # find (x - head2.data) in 'us'
        if ((x - head2.data) in us):
            count += 1
          
        # move to next node
        head2 = head2.next;   
     
    # required count of pairs    
    return count;
  
# Driver program to test above
if __name__=='__main__':
     
    head1 = None;
    head2 = None;
      
    # create linked list1 3.1.5.7
    head1 = push(head1, 7);
    head1 = push(head1, 5);
    head1 = push(head1, 1);
    head1 = push(head1, 3);   
      
    # create linked list2 8.2.5.3
    head2 = push(head2, 3);
    head2 = push(head2, 5);
    head2 = push(head2, 2);
    head2 = push(head2, 8);
      
    x = 10;
      
    print("Count =", countPairs(head1, head2, x));
     
# This code is contributed by rutvik_56

C＃

// C# implementation to count pairs from both linked
// lists whose sum is equal to a given value
 
// Note : here we use java.util.LinkedList for
// linked list implementation
using System;
using System.Collections.Generic;
 
class GFG
{
    // method to count all pairs from both the linked lists
    // whose sum is equal to a given value
    static int countPairs(List head1, List head2, int x)
    {
        int count = 0;
         
        HashSet us = new HashSet();
         
        // insert all the elements of 1st list
        // in the hash table(unordered_set 'us')
        foreach(int itr1 in head1)
        {
            us.Add(itr1);
             
        }
 
        // for each element of 2nd list
        foreach(int itr2 in head2)
        {
            // find (x - head2->data) in 'us'
            if (!(us.Contains(x - itr2)))
                count++;
                 
        }
         
        // required count of pairs    
        return count;
    }
     
    // Driver code
    public static void Main(String[] args)
    {
        int []arr1 = {3, 1, 5, 7};
        int []arr2 = {8, 2, 5, 3};
         
        // create linked list1 3->1->5->7
        List head1 = new List(arr1);
         
        // create linked list2 8->2->5->3
        List head2 = new List(arr2);
         
        int x = 10;
         
        Console.WriteLine("Count = " + countPairs(head1, head2, x));
    }
}
 
// This code is contributed by Princi Singh

输出：

Count = 2

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