在数字列表中找到最小数字的递归方法 [英] Recursive method to find the minimum number in a list of numbers

问题描述

鉴于此示例列表:

[5, 3, 9, 10, 8, 2, 7]

如何使用递归求最小数?答案是2.

How to find the minimum number using recursion? The answer is 2.

我在做递归练习时在试卷中发现了这一点.我想不出解决这个问题的方法.要找到它，我是否必须先对列表进行排序，然后递归地无事可做.谁能告诉我路径?

I found this in a question paper while I was doing recursion exercises. I can't figure out a way to solve this. To find it, do I have to sort the list first and then there's nothing to do recursively. Can any one show me a path?

推荐答案

这是min的递归实现:

l=[5, 3, 9, 10, 8, 2, 7]
def find_min(l,current_minimum = None):
    if not l:
        return current_minimum
    candidate=l.pop()
    if current_minimum==None or candidate<current_minimum:
        return find_min(l,candidate)
    return find_min(l,current_minimum)
print find_min(l)
>>>
2     

考虑到这不应在实际程序中使用，而应视为练习.性能会比内置的min差几个数量级.

Take into account that this should not be used in real programs and should be treated as an exercise. The performance will be worse than the built-in minby several orders of magnitude.

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