Given a singly linked list and a key, count number of occurrences of given key in linked list. For example, if given linked list is 1->2->1->2->1->3->1 and given key is 1, then output should be 4.

Algorithm:

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.

Python Programming:

# 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 occurances of a node
    # (seach_for) in a linkded 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)

Output:

count of 1 is 3

Time Complexity: O(n)
Auxiliary Space: O(1)

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