为什么此代码不返回反向链接列表

Question

Input = [1, 2, 3, 4, 5]

    def reverseList(self, head):
    prev = None
    while head:
        cur = head
        cur.next = prev
        head = head.next
        prev = cur

    return prev

I do not understand why the result above returns

Output = [1]

cur is temporary variable and not affected by changes of the head variable (that is what I thought)

    def reverseList(self, head):
    prev = None
    while head:
        cur = head
        head = head.next
        cur.next = prev
        prev = cur

    return prev

returns the correct result, and I do not understand why. It suggests that changes to head also change the cur variable, although it is changed AFTER cur is set to cur = head

Answer 1

This is your example linked list:

[1] -> [2] -> [3] -> [4] -> [5]

Initially head is pointing to [1] and prev is pointing to None . Now let's execute your first code line by line. You first set cur to head (eg, cur is now pointing to [1] ), meaning from now on both head and cur is pointing to [1] . Now, when you set cur.next = prev you are actually setting [1].next = None . So, the link between [1] and [2] is already broken and you loose all the other parts of the linked-list (eg, [2] -> [3] -> [4] -> [5] ) at this point. Now when you do head = head.next , this means you are doing [1] = [1].next , where previously you set [1].next = None . In the next run, the while loop will terminate because the head become None .

I would highly suggest you to do this type of simulation using pen and paper for the second code and you will understand the difference.