Explanation
In this program, we will create a doubly linked list and count the number of nodes present in the list. To count the node, we traverse through the list by incrementing the counter by 1.
What count of nodes presents above doubly linked list is 5.
Algorithm
- Define a Node class which represents a node in the list. It will have three properties: data, previous which will point to the previous node and next which will point to the next node.
- Define another class for creating a doubly linked list, and it has two nodes: head and tail. Initially, head and tail will point to null.
- addNode() will add node to the list:
- It first checks whether the head is null, then it will insert the node as the head.
- Both head and tail will point to a newly added node.
- Head's previous pointer will point to null and tail?s next pointer will point to null.
- If the head is not null, the new node will be inserted at the end of the list such that new node's previous pointer will point to tail.
- The new node will become the new tail. Tail's next pointer will point to null.
- countNodes() will count the number of nodes present in the list.
- Define a variable counter and new node current which will point to the head node.
- Traverse through the list to count the nodes by making the current node to point to next node in the list till current point to null.
- Increment the counter by 1.
- display() will show all the nodes present in the list.
- Define a new node 'current' that will point to the head.
- Print current.data till current points to null.
- Current will point to the next node in the list in each iteration.
Input:
#Add nodes to the list
dList.addNode(1);
dList.addNode(2);
dList.addNode(3);
dList.addNode(4);
dList.addNode(5);
Output:
Nodes of doubly linked list: 1 2 3 4 5
Count of nodes present in the list: 5
Python
Output:
C
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JAVA
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C#
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PHP
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