CODEWITHSUNDEEP
pythonlinked-list
Pxx
Pxx

Reputation: 11

Linked List problem dummy.next different different as head

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution(object):
    def deleteDuplicates(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        dummy = ListNode()
        dummy.next = head
        while dummy and dummy.next:
            if dummy.val == dummy.next.val:
                dummy.next = dummy.next.next
            else:
                dummy = dummy.next
        return dummy.next

So I'm doing one of the Leetcode problem call remove duplicate from sorted list, i created a dummy node which it's next is head, however when i try to test it with head = [1,2,3,3,4,4,5], return dummy.next is giving a empty list as output and return head is giving the correct output, i'm wondering why is that since dummy.next should be equal to head?

Upvotes: 1

Views: 125

Answers (1)

trincot
trincot

Reputation: 350310

The idea to use this dummy is good, but then you shouldn't assign anything else to it. In your loop you should use another name for traversing the list -- not dummy: this could be a new name like curr, or you could even use head for it:

        dummy = ListNode()
        dummy.next = head
        while head and head.next:
            if head.val == head.next.val:
                head.next = head.next.next
            else:
                head = head.next
        return dummy.next

Upvotes: 3

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