合并两个列表并递归排序 [英] combine two list and sort recursively

问题描述

我正在尝试定义一个名为 combine 的函数，以将两个列表作为参数并递归地将它们组合成一个排序列表.

I'm trying to define a function named combine to take two lists as parameters and combine them into one sorted list recursively.

def combine (l1,l2):
    if l1 == []:
        return l2
    elif l2 == []:
        return l1
    elif l1 == [] and l2 == []:
        return []
    sort_l = []
    index = 0
    for num in l1:
        if num <= l2[0]:
            sort_l.append(num)
            index += 1
        else:
            return combine(l2, l1[index:])
    return sort_l + l2

它给了我:

combine([1,3,5,7,9],[0,2,4,6,8]) -> [8, 9] 

但应该:

[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

有什么建议吗???

推荐答案

你的方法是正确的.但是，您要在每次递归调用中重新分配 sort_l.您需要在函数外部定义它或在每次函数调用时传递它，只创建一次:

Your approach is correct. But, you are reallocating the sort_l in every recursive call. You need to either define it outside the function or pass it with every function call, after creating only once:

sort_l = []

def combine (l1,l2):
    if l1 == []:
        return l2
    elif l2 == []:
        return l1
    elif l1 == [] and l2 == []:
        return []
    index = 0
    for num in l1:
        if num <= l2[0]:
            sort_l.append(num)
            index += 1
        else:
            return combine(l2, l1[index:]) 
    return sort_l + l2

或

def combine (l1,l2, sort_l=None):
    if sort_l is None:
       sort_l = []
    if l1 == []:
        return l2
    elif l2 == []:
        return l1
    elif l1 == [] and l2 == []:
        return []
    index = 0
    for num in l1:
        if num <= l2[0]:
            sort_l.append(num)
            index += 1
        else:
            return combine(l2, l1[index:], sort_l) # pass it here
    return sort_l + l2

combine([1,3,5,7,9],[0,2,4,6,8]) # Don't pass any list, it will be created in first call.

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