在不规则列表中查找最长列表的长度 [英] Finding length of the longest list in an irregular list of lists

问题描述

我必须在列表列表中找到最长的列表.

I have to find the longest list inside a list of lists.

例如:

longest([1,2,3])返回3

longest([[[[1,2,3]]])也返回3(内部列表为3)

longest([[[1,2,3]]]) also returns 3 (inner list is 3)

longest([[]，[3，[4,5]，[2,3,4,5,3,3]，[7]，5，[1,2,3]，[3,4]]，[1,2,3,4,5]])返回7(列表 [3，[4,5]，[2,3,4,5,3，3]，[7]，5，[1,2,3]，[3,4]] 包含7个元素)

longest([[], [3,[4,5],[2,3,4,5,3,3], [7], 5, [1,2,3], [3,4]], [1,2,3,4,5]]) returns 7 (list [3,[4,5],[2,3,4,5,3,3], [7], 5, [1,2,3], [3,4]] contains 7 elements)

现在我有了这段代码，但是前两个示例并没有解决问题.

Right now I have this code, but it doesn't do the trick with the first two examples.

def longest(list1):
    longest_list = max(len(elem) for elem in list1)
    return longest_list

也许递归会有所帮助?

推荐答案

以下是任何深度列表的递归解决方案:

Here is a recursive solution for any depth list:

def longest(l):
    if not isinstance(l, list):
        return 0
    return max(
            [len(l)] 
            + [len(subl) for subl in l if isinstance(subl, list)] 
            + [longest(subl) for subl in l]
            )

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