编写一个函数，计算一个给定的int在链表中出现的次数

本文概述

• C ++
• C
• Java
• python
• C#
• C ++
• Java
• Python3
• C#
• C ++
• Java
• C#
• Python3

给定一个单链表和一个键, 计算给定键在链表中出现的次数。例如, 如果给定的链表为1-> 2-> 1-> 2-> 1-> 3-> 1, 且给定键为1, 则输出应为4。

方法1-无递归

算法：

1. Initialize count as zero.
2. Loop through each element of linked list:
     a) If element data is equal to the passed number then
        increment the count.
3. Return count.

实现

C ++

// C++ program to count occurrences in a linked list
#include <bits/stdc++.h>
using namespace std;
  
/* Link list node */
class Node {
public :
     int data;
     Node* next;
};
  
/* Given a reference (pointer to pointer) to the head 
of a list and an int, push a new node on the front 
of the list. */
void push(Node** head_ref, int new_data)
{
     /* allocate node */
     Node* new_node = new 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;
}
  
/* Counts the no. of occurrences of a node 
(search_for) in a linked list (head)*/
int count(Node* head, int search_for)
{
     Node* current = head;
     int count = 0;
     while (current != NULL) {
         if (current->data == search_for)
             count++;
         current = current->next;
     }
     return count;
}
  
/* Driver program to test count function*/
int main()
{
     /* Start with the empty list */
     Node* head = NULL;
  
     /* Use push() to construct below list 
     1->2->1->3->1 */
     push(&head, 1);
     push(&head, 3);
     push(&head, 1);
     push(&head, 2);
     push(&head, 1);
  
     /* Check the count function */
     cout << "count of 1 is " << count(head, 1);
     return 0;
}
  
// This is code is contributed by rathbhupendra

C

// C program to count occurrences in a linked list
#include <stdio.h>
#include <stdlib.h>
  
/* Link list node */
struct Node {
     int data;
     struct Node* next;
};
  
/* Given a reference (pointer to pointer) to the head
   of a list and an int, push a new node on the front
   of the 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;
}
  
/* Counts the no. of occurrences of a node
    (search_for) in a linked list (head)*/
int count( struct Node* head, int search_for)
{
     struct Node* current = head;
     int count = 0;
     while (current != NULL) {
         if (current->data == search_for)
             count++;
         current = current->next;
     }
     return count;
}
  
/* Driver program to test count function*/
int main()
{
     /* Start with the empty list */
     struct Node* head = NULL;
  
     /* Use push() to construct below list
      1->2->1->3->1  */
     push(&head, 1);
     push(&head, 3);
     push(&head, 1);
     push(&head, 2);
     push(&head, 1);
  
     /* Check the count function */
     printf ( "count of 1 is %d" , count(head, 1));
     return 0;
}

Java

// Java program to count occurrences in a linked list
class LinkedList {
     Node head; // head of list
  
     /* Linked list Node*/
     class Node {
         int data;
         Node next;
         Node( int d)
         {
             data = d;
             next = null ;
         }
     }
  
     /* Inserts a new Node at front of the list. */
     public 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;
     }
  
     /* Counts the no. of occurrences of a node
     (search_for) in a linked list (head)*/
     int count( int search_for)
     {
         Node current = head;
         int count = 0 ;
         while (current != null ) {
             if (current.data == search_for)
                 count++;
             current = current.next;
         }
         return count;
     }
  
     /* Driver function to test the above methods */
     public static void main(String args[])
     {
         LinkedList llist = new LinkedList();
  
         /* Use push() to construct below list
           1->2->1->3->1  */
         llist.push( 1 );
         llist.push( 2 );
         llist.push( 1 );
         llist.push( 3 );
         llist.push( 1 );
  
         /*Checking count function*/
         System.out.println( "Count of 1 is " + llist.count( 1 ));
     }
}
// This code is contributed by Rajat Mishra

python

# Python program to count the number of time a given
# int occurs in a linked list
  
# Node class 
class Node:
  
     # Constructor to initialize the node object
     def __init__( self , data):
         self .data = data
         self . next = None
  
class LinkedList:
  
     # Function to initialize head
     def __init__( self ):
         self .head = None
  
     # Counts the no . of occurrences of a node
     # (search_for) in a linked list (head)
     def count( self , search_for):
         current = self .head
         count = 0
         while (current is not None ):
             if current.data = = search_for:
                 count + = 1
             current = current. next
         return count
  
     # Function to insert a new node at the beginning
     def push( self , new_data):
         new_node = Node(new_data)
         new_node. next = self .head
         self .head = new_node
  
     # Utility function to print the linked LinkedList
     def printList( self ):
         temp = self .head
         while (temp):
             print temp.data, temp = temp. next
  
  
# Driver program
llist = LinkedList()
llist.push( 1 )
llist.push( 3 )
llist.push( 1 )
llist.push( 2 )
llist.push( 1 )
  
# Check for the count function
print "count of 1 is % d" % (llist.count( 1 ))
  
# This code is contributed by Nikhil Kumar Singh(nickzuck_007)

C#

// C# program to count occurrences in a linked list
using System;
class LinkedList {
     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 ;
         }
     }
  
     /* Inserts a new Node at front of the list. */
     public 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;
     }
  
     /* Counts the no. of occurrences of a node 
     (search_for) in a linked list (head)*/
     int count( int search_for)
     {
         Node current = head;
         int count = 0;
         while (current != null ) {
             if (current.data == search_for)
                 count++;
             current = current.next;
         }
         return count;
     }
  
     /* Driver code */
     public static void Main(String[] args)
     {
         LinkedList llist = new LinkedList();
  
         /* Use push() to construct below list 
         1->2->1->3->1 */
         llist.push(1);
         llist.push(2);
         llist.push(1);
         llist.push(3);
         llist.push(1);
  
         /*Checking count function*/
         Console.WriteLine( "Count of 1 is " + llist.count(1));
     }
}
  
// This code is contributed by Arnab Kundu

输出如下：

count of 1 is 3

时间复杂度：

O(n)

辅助空间：

O(1)

方法2-递归

该方法是由

MY_DOOM

.

算法：

Algorithm
count(head, key);
if head is NULL
return frequency
if(head->data==key)
increase frequency by 1
count(head->next, key)

实现

C ++

// C++ program to count occurrences in
// a linked list using recursion
#include <bits/stdc++.h>
using namespace std;
  
/* Link list node */
struct Node {
     int data;
     struct Node* next;
};
// global variable for counting frequeancy of
// given element k
int frequency = 0;
  
/* Given a reference (pointer to pointer) to the head
of a list and an int, push a new node on the front
of the 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;
}
  
/* Counts the no. of occurrences of a node
(search_for) in a linked list (head)*/
int count( struct Node* head, int key)
{
     if (head == NULL)
         return frequency;
     if (head->data == key)
         frequency++;
     return count(head->next, key);
}
  
/* Driver program to test count function*/
int main()
{
     /* Start with the empty list */
     struct Node* head = NULL;
  
     /* Use push() to construct below list
      1->2->1->3->1  */
     push(&head, 1);
     push(&head, 3);
     push(&head, 1);
     push(&head, 2);
     push(&head, 1);
  
     /* Check the count function */
     cout << "count of 1 is " << count(head, 1);
     return 0;
}

Java

// Java program to count occurrences in
// a linked list using recursion
import java.io.*;
import java.util.*;
  
// Represents node of a linkedlist
class Node {
     int data;
     Node next;
     Node( int val)
     {
         data = val;
         next = null ;
     }
}
  
class GFG {
  
     // global variable for counting frequeancy of
     // given element k
     static int frequency = 0 ;
  
     /* Given a reference (pointer to pointer) to the head 
     of a list and an int, push a new node on the front 
     of the list. */
  
     static Node push(Node head, int new_data)
     {
         // allocate node
         Node new_node = new Node(new_data);
  
         // link the old list off the new node
         new_node.next = head;
  
         // move the head to point to the new node
         head = new_node;
  
         return head;
     }
  
     /* Counts the no. of occurrences of a node 
     (search_for) in a linked list (head)*/
     static int count(Node head, int key)
     {
         if (head == null )
             return frequency;
         if (head.data == key)
             frequency++;
         return count(head.next, key);
     }
  
     // Driver Code
     public static void main(String args[])
     {
         // Start with the empty list
         Node head = null ;
  
         /* Use push() to construct below list 
         1->2->1->3->1 */
         head = push(head, 1 );
         head = push(head, 3 );
         head = push(head, 1 );
         head = push(head, 2 );
         head = push(head, 1 );
  
         /* Check the count function */
         System.out.print( "count of 1 is " + count(head, 1 ));
     }
}
  
// This code is contributed by rachana soma

Python3

# Python program to count the number of 
# time a given int occurs in a linked list
# Node class
class Node:
      
     # Constructor to initialize the node object
     def __init__( self , data):
         self .data = data
         self . next = None
  
class LinkedList:
      
     # Function to initialize head
     def __init__( self ):
         self .head = None
         self .counter = 0
          
     # Counts the no . of occurances of a node
     # (seach_for) in a linkded list (head)
     def count( self , li, key):     
          
         # Base case 
         if ( not li): 
             return self .counter
          
         # If key is present in 
         # current node, return true 
         if (li.data = = key): 
             self .counter = self .counter + 1
          
         # Recur for remaining list 
         return self .count(li. next , key) 
  
     # Function to insert a new node 
     # at the beginning
     def push( self , new_data):
         new_node = Node(new_data)
         new_node. next = self .head
         self .head = new_node
  
     # Utility function to print the 
     # linked LinkedList
     def printList( self ):
         temp = self .head
         while (temp):
             print (temp.data)
             temp = temp. next
  
# Driver Code
llist = LinkedList()
llist.push( 1 )
llist.push( 3 )
llist.push( 1 )
llist.push( 2 )
llist.push( 1 )
  
# Check for the count function
print ( "count of 1 is" , llist.count(llist.head, 1 ))
  
# This code is contributed by 
# Gaurav Kumar Raghav

C#

// C# program to count occurrences in
// a linked list using recursion
using System;
  
// Represents node of a linkedlist
public class Node {
     public int data;
     public Node next;
     public Node( int val)
     {
         data = val;
         next = null ;
     }
}
  
class GFG {
  
     // global variable for counting frequeancy of
     // given element k
     static int frequency = 0;
  
     /* Given a reference (pointer to pointer) to the head 
     of a list and an int, push a new node on the front 
     of the list. */
  
     static Node push(Node head, int new_data)
     {
         // allocate node
         Node new_node = new Node(new_data);
  
         // link the old list off the new node
         new_node.next = head;
  
         // move the head to point to the new node
         head = new_node;
  
         return head;
     }
  
     /* Counts the no. of occurrences of a node 
     (search_for) in a linked list (head)*/
     static int count(Node head, int key)
     {
         if (head == null )
             return frequency;
         if (head.data == key)
             frequency++;
         return count(head.next, key);
     }
  
     // Driver Code
     public static void Main(String[] args)
     {
         // Start with the empty list
         Node head = null ;
  
         /* Use push() to construct below list 
         1->2->1->3->1 */
         head = push(head, 1);
         head = push(head, 3);
         head = push(head, 1);
         head = push(head, 2);
         head = push(head, 1);
  
         /* Check the count function */
         Console.Write( "count of 1 is " + count(head, 1));
     }
}
  
/* This code contributed by PrinciRaj1992 */

输出如下：

count of 1 is 3

下面的方法可用于避免全局变量" frequency"(在Python 3代码的情况下为反计数)。

C ++

// method can be used to avoid
// Global variable 'frequency'
  
/* Counts the no. of occurrences of a node 
(search_for) in a linked list (head)*/
int count( struct Node* head, int key)
{
     if (head == NULL)
         return 0;
     if (head->data == key)
         return 1 + count(head->next, key);
     return count(head->next, key);
}

Java

// method can be used to avoid
// Global variable 'frequency'
  
/* Counts the no. of occurrences of a node 
(search_for) in a linked list (head)*/
int count(Node head, int key)
{
     if (head == null )
         return 0 ;
     if (head.data == key)
         return 1 + count(head.next, key);
     return count(head.next, key);
}
  
// This code is contributed by rachana soma

C#

// method can be used to avoid
// Global variable 'frequency'
using System; 
  
/* Counts the no. of occurrences of a node 
(search_for) in a linked list (head)*/
static int count(Node head, int key) 
{ 
     if (head == null ) 
         return 0; 
     if (head.data == key) 
         return 1 + count(head.next, key);
     return count(head.next, key); 
} 
  
// This code is contributed by SHUBHAMSINGH10

Python3

def count( self , temp, key):
      
     # during the initial call, temp
     # has the value of head
      
     # Base case
     if temp is None :
         return 0
          
     # if a match is found, we 
     # increment the counter
     if temp.data = = key:
         return 1 + count(temp. next , key)
     return count(temp. next , key)
      
# to call count, use
# linked_list_name.count(head, key)

以上方法实现了头递归。下面给出了相同的尾部递归实现。谢谢浦那·Ja那建议这种方法：

int count(struct Node* head, int key)
{
    if(head == NULL)
        return 0;
        
   int frequency = count(head->next, key);
   if(head->data == key)
     return 1 + frequency;
    
    // else 
    return frequency;
}

时间复杂度：O(n)

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