编写一个函数来计算给定的 int 在链表中出现的次数

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，如果给定的链表是 1->2->1->2->1->3->1 并且给定的键是 1，那么输出应该是 4。

方法 1- 不使用递归

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. 

执行：

C++

// C++ program to count occurrences in a linked list
#include 
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 
#include 
 
/* 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

Javascript

// javascript program to count occurrences in a linked list
var head; // head of list
 
    /* Linked list Node */
      
     class Node {
            constructor(val) {
                this.data = val;
                this.next = null;
            }
        }
 
    /* Inserts a new Node at front of the list. */
    function push(new_data) {
        /*
         * 1 & 2: Allocate the Node & Put in the data
         */
         new_node = new Node(new_data);
 
        /* 3. Make next of new Node as head */
        new_node.next = head;
 
        /* 4. Move the head to povar to new Node */
        head = new_node;
    }
 
    /*
     * Counts the no. of occurrences of a node (search_for) in a linked list (head)
     */
    function count(search_for) {
         current = head;
        var count = 0;
        while (current != null) {
            if (current.data == search_for)
                count++;
            current = current.next;
        }
        return count;
    }
 
    /* Driver function to test the above methods */
     
         
 
        /*
         * Use push() to construct below list 1->2->1->3->1
         */
        push(1);
        push(2);
        push(1);
        push(3);
        push(1);
 
        /* Checking count function */
        document.write("Count of 1 is " + count(1));
 
// This code contributed by gauravrajput1

输出：

count of 1 is 3

时间复杂度： O(n)
辅助空间： O(1)

方法 2- 使用递归
此方法由 MY_DOOM 贡献。

算法：

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

执行：

C++

// C++ program to count occurrences in
// a linked list using recursion
#include 
using namespace std;
 
/* Link list node */
struct Node {
    int data;
    struct Node* next;
};
// global variable for counting frequency 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 frequency 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

蟒蛇3

# 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 occurrences 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 frequency 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 */

Javascript

 
// Javascript program to count occurrences in
// a linked list using recursion
 
// Represents node of a linkedlist
class Node
{
    constructor(val)
    {
        this.data = val;
        this.next = null;
    }
}
 
// Global variable for counting
// frequency of given element k
let 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. */
function push(head, new_data)
{
     
    // Allocate node
    let 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)*/
function count(head, key)
{
    if (head == null)
        return frequency;
    if (head.data == key)
        frequency++;
         
    return count(head.next, key);
}
 
// Driver code
 
// Start with the empty list
let 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 */
document.write("count of 1 is " +
               count(head, 1));
                
// This code is contributed by divyeshrabadiya07
 

输出：

count of 1 is 3

下面的方法可用于避免全局变量“频率”（ Python 3 代码中的计数器）。

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

蟒蛇3

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)

Javascript

 
// method can be used to afunction
// Global variable 'frequency'
 
/* Counts the no. of occurrences of a node
(search_for) in a linked list (head)*/
function count(head , 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 todaysgaurav
 

上述方法实现了头递归。下面给出的是相同的尾递归实现。感谢Puneet Jain提出这种方法：

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)

https://www.youtube.com/watch?v=-g3KauAWofw

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