将两个已排序的链接列表合并到python中的一个链接列表中

here is my code:

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

the problem here is stucked in the infinite loop.can any one tell me what the problem is

这里的问题被困在无限循环中。任何人都告诉我问题是什么

the examples are:

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])

2 个解决方案

#1

The problem with the current code is that it causes a side-effect of the temp node's next before it navigates to the next node from the current node. This is problematic when the current temp node is the current node.

当前代码的问题在于，在导航到当前节点的下一个节点之前，它会导致临时节点的副作用。当前临时节点是当前节点时，这是有问题的。

That is, imagine this case:

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

There is a corrected version, with some other updates:

有一个更正版本，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

#2

Recursive algorithm for merging two sorted linked lists

用于合并两个已排序链接列表的递归算法

def merge_lists(h1, h2):
    if h1 is None:
        return h2
    if h2 is None:
        return h1

    if (h1.value < h2.value):
        h1.next = merge_lists(h1.next, h2)
        return h1
    else:
        h2.next = merge_lists(h2.next, h1)
        return h2

#1