python - 在python中将两个排序的链表合并为一个链表

Question

这是我的代码:

def merge_lists(head1, head2):
    if head1 is None and head2 is None:
        return None
    if head1 is None:
        return head2
    if head2 is None:
        return head1
    if head1.value < head2.value:
        temp = head1
    else:
        temp = head2
    while head1 != None and head2 != None:
        if head1.value < head2.value:
            temp.next = head1
            head1 = head1.next
        else:
            temp.next = head2
            head2 = head2.next
    if head1 is None:
        temp.next = head2
    else:
        temp.next = head1
    return temp
    pass

这里的问题卡在死循环了，谁能告诉我是什么问题

例子是:

 assert [] == merge_lists([],[])
 assert [1,2,3] == merge_lists([1,2,3], [])
 assert [1,2,3] == merge_lists([], [1,2,3])
 assert [1,1,2,2,3,3,4,5] == merge_lists([1,2,3], [1,2,3,4,5])

Answer 1

当前代码的问题在于它会导致临时节点的下一个之前从当前节点导航到下一个节点的副作用。当当前临时节点是当前节点时，这是有问题的。

也就是说，想象一下这种情况:

temp = N
temp.next = N  # which means N.next = N
N = N.next     # but from above N = (N.next = N) -> N = N

有一个更正的版本，还有一些其他更新:

def merge_lists(head1, head2):
    if head1 is None:
        return head2
    if head2 is None:
        return head1

    # create dummy node to avoid additional checks in loop
    s = t = node() 
    while not (head1 is None or head2 is None):
        if head1.value < head2.value:
            # remember current low-node
            c = head1
            # follow ->next
            head1 = head1.next
        else:
            # remember current low-node
            c = head2
            # follow ->next
            head2 = head2.next

        # only mutate the node AFTER we have followed ->next
        t.next = c          
        # and make sure we also advance the temp
        t = t.next

    t.next = head1 or head2

    # return tail of dummy node
    return s.next