Python Linked List removal element

Question

My code is like:

class ListNode(object):
    def __init__(self, x):
        self.val = x
        self.next = None

class Solution:
    # @param {ListNode} head
    # @param {integer} val
    # @return {ListNode}
    def removeElements(self, head, val):
        def printLinked(head):
            while head:
                print head.val,"->"
                head = head.next

        dummy = ListNode(0)
        dummy.next = head
        cur = dummy
        while cur and cur.next:
            if cur.next.val == val:
                cur.next = cur.next.next
            else:
                cur = cur.next
        printLinked(dummy.next)
        print "._."
        printLinked(head)
        return head

I find the if I return dummy.next rather than head, my code will be correct. But, I cannot understand what's the difference between head and dummy.next.

When my test case is head = ListNode(1),val = 1. The head will still be 1, but the dummy.next is [].

I am quite confused, because when the test case is head = 1 --> 2 --> 6 --> 3 --> 4 --> 5 --> 6, val = 6, both head and dummy.next is 1-->2-->3-->4-->5. And I think there must be some misunderstand of head and dummy for me. Can any one explain it for me?

Answer 1

Based on Kendas's comment, "The cur.next attribute points to the same object head points to. When you give cur.next another value (cur.next == cur.next.next), it doesn't point to that object anymore. However, the variable head still points to the same object it did before." Hope this helps.