Jason Piao - 6 months ago 34

Python Question

Im doing a pretty simple leetcode question and got the answer, however, upon further inspection, I found something that doesn't make all that much sense to me.

The question:

Given a linked list, swap every two adjacent nodes and return its

head.

For example, Given 1->2->3->4, you should return the list as

2->1->4->3.

Your algorithm should use only constant space. You may not modify the

values in the list, only nodes itself can be changed.

My solution:

`# Definition for singly-linked list.`

# class ListNode(object):

# def __init__(self, x):

# self.val = x

# self.next = None

class Solution(object):

def swapPairs(self, head):

"""

:type head: ListNode

:rtype: ListNode

"""

dummy = ListNode(0)

dummy.next = head

curr = dummy

while curr.next and curr.next.next:

a = curr.next

b = curr.next.next

curr.next = b

a.next = b.next

b.next = a

curr = a

return dummy.next

The confusion:

When I do

`curr = dummy`

`curr`

`dummy`

`return dummy.next`

`return curr.next`

`curr = dummy`

`curr`

`dummy`

`dummy`

`dummy.next`

Also, when I do:

`if curr is dummy:`

print("SS")

while curr.next and curr.next.next:

a = curr.next

b = curr.next.next

curr.next = b

a.next = b.next

b.next = a

curr = a

if curr is dummy:

print("AAA")

`SS`

`AAA`

Can someone clear the confusion up for me?

Thanks!

Recommended for you: Get network issues from **WhatsUp Gold**. **Not end users.**

Answer Source

When you do an assignment `var = x`

you make `var`

point to `x`

. It wouldn't make other variables pointed to the same object point to `x`

after this.

Consider this example:

```
>>> a = {'p': 2}
>>> b = a
>>> b
{'p': 2}
>>> b['p'] = 4
>>> a
{'p': 4}
>>> b = {'q': 3}
>>> a
{'p': 4}
>>> b
{'q': 3}
```

Or this:

```
>>> class Node:
... def __init__(self, nxt):
... self.nxt = nxt
... def __repr__(self):
... return '(Node => %s)' % self.nxt
...
>>> a = Node(Node(2))
>>> a
(Node => (Node => 2))
>>> b = a
>>> b
(Node => (Node => 2))
>>> a.nxt = Node(4)
>>> a
(Node => (Node => 4))
>>> b
(Node => (Node => 4))
>>> b = b.nxt
>>> a
(Node => (Node => 4))
>>> b
(Node => 4)
```

Recommended from our users: **Dynamic Network Monitoring from WhatsUp Gold from IPSwitch**. ** Free Download**