带有两个迭代器的python一次递增1

我是python的新手.

我尝试在python中实现诸如归类排序之类的操作.我使用https://pythonhosted.org/llist/#sllist-objects的单链列表.要合并两个排序的列表,我需要使用迭代器遍历两个列表.伪代码如下所示：

n3 = sllist()
for n1 in list1 and n2 in list2:
    if (n1 > n2):
        n3.append(n1)
        n1++               # c way of doing thing
    else:
        n3.append(n2)
        n2++               # c way of doing thing

但是我不知道如何使它在python中工作.任何指针或提示都会有所帮助.

编辑：
在所有讨论和建议之后,我想到了类似这样的代码.谁能告诉我,如何摆脱最后两个while循环.我打算使用“扩展”,但无法使用它.

final_list = sllist()
node1 = list1.first
node2 = list2.first

while node1 and node2:
    if node1.value < node2.value:
        final_list.append(node1)
        node1 = node1.next              
    else:
        final_list.append(node2)
        node2 = node2.next
while node1:
    final_list.append(node1)
    node1 = node1.next
while node2:
    final_list.append(node2)
    node2 = node2.next

return final_list

解决方法:

我通常使用iterables和next执行此操作：

lst1 = iter(list1)
lst2 = iter(list2)
out = sllist()
sentinel = object()
n1 = next(lst1, sentinel)
n2 = next(lst2, sentinel)
while n1 is not sentinel and n2 is not sentinel:
    if n1 > n2:
        out.append(n2)
        n2 = next(lst2, sentinel)
    elif n2 >= n1:
        out.append(n1)
        n1 = next(lst1, sentinel)
out.extend(lst1)
out.extend(lst2)

正如评论中指出的那样,您还可以将其编写为：

lst1 = iter(list1)
lst2 = iter(list2)
out = sllist()
try:
    n1 = next(lst1)
    n2 = next(lst2)
    while True:
        if n1 > n2:
            out.append(n2)
            n2 = next(lst2)
        elif n2 >= n1:
            out.append(n1)
            n1 = next(lst1)

except StopIteration:  # raised when next(...) fails.
    out.extend(lst1)
    out.extend(lst2)

它在功能上是等效的.选择您喜欢的任何一个:-)

标签：mergesort,iterator,linked-list,python
来源： https://codeday.me/bug/20191029/1958319.html