python - Python 中的引用和值传递

Question

我正在做一道 LeetCode 的题。我发现了一些对我来说没有多大意义的东西。

问题:

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.

我的解决方案:

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

困惑:

当我执行 curr = dummy 时，似乎 dummy 是通过引用传递的，并且变异 curr 会变异 dummy。但是return语句return dummy.next和return curr.next的值不一样？由于 curr = dummy，在循环结束时，curr 位于列表的末尾，因此 dummy 不应该位于 list 的末尾？但是，dummy 仍然位于列表的最前面，dummy.next 是传递到函数中的列表的开头，但以正确的方式进行了突变。

另外，当我这样做时:

        if curr is dummy:
            print("BEGIN")

        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("END")

BEGIN 被打印，但 END 没有打印。

有人可以帮我解答疑惑吗？

Answer 1

当您进行赋值时，var = x 使var 指向x。在此之后，它不会使指向同一对象的其他变量指向 x。

考虑这个例子:

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

或者这个:

>>> 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)