Count Linked List Nodes

Question

I am trying to count the numbers of nodes in a linked list when the node value is a non-negative odd, but it seems that I couldn't get the right result.

class Solution:
    """
    @param head: 
    @return: nothing
    """
    def countNodesII(self, head):
        count = 0
        while head.next is not None:
            head = head.next
            if head.val > 0 and head.val % 2 != 0:
                count += 1 
            else:
                return 0
        return count

if the input is 1->3->5->null I expect to get result of 3, but my program returned 2 instead.

Devesh Kumar Singh · Accepted Answer · 2019-06-05 09:14:22Z

2

Some issues in your code

You would want to move to next node via head=head.next after you check the head node for the condition, right now you are skipping head
You would want to remove return from else, and only return count at the end
You want to check if head reaches None since you are using it to iterate over your list

So the updated code will look like

class Solution:
    """
    @param head: 
    @return: nothing
    """
    def countNodesII(self, head):
        count = 0
        #Iterate over the list
        while head != None:
            #Check for condition
            if head.val > 0 and head.val % 2 != 0:
                count += 1
            #Move to next node
            head = head.next
        return count

answered Jun 5, 2019 at 9:14

Devesh Kumar Singh

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3 Comments

Netwave Over a year ago

Exactly this :D

Netwave Over a year ago

That this is the exact and simplest solution ;)

jason Over a year ago

Thanks, it's really helpful!

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