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 = https://www.lsbin.com/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 = https://www.lsbin.com/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 = https://www.lsbin.com/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 = https://www.lsbin.com/data self . next = Noneclass 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 = https://www.lsbin.com/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=https://www.lsbin.com/=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 = https://www.lsbin.com/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 = https://www.lsbin.com/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 = https://www.lsbin.com/data self . next = Noneclass 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 = https://www.lsbin.com/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 =https://www.lsbin.com/= 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 =https://www.lsbin.com/= 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 =https://www.lsbin.com/= 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 = https://www.lsbin.com/= 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 =https://www.lsbin.com/= key) return 1 + frequency; // else return frequency; }

时间复杂度：O(n)
【编写一个函数，计算一个给定的int在链表中出现的次数】如果发现任何不正确的地方, 或者想分享有关上述主题的更多信息, 请写评论。

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