用递归查找所有索引 [英] Find all indices with recursion

问题描述

我必须编写一个程序来查找列表或字符串中特定元素的所有索引值.我必须使用递归，而我的函数只能接受两个参数.

I have to write a program that finds all of the index values for specific element in list or a string. I have to use recursion, and my function can only take two arguments.

我的问题是我的程序只找到第一个索引然后停止.如何更改它以满足我的要求?

My problem is that my program finds only the first index and then stop. How can I change it to meet my requirements?

我的代码:

def find_all(L, v):
    return 0 if L[0] == v else 1 + find_all(L[1:], v)

输入:

1. find_all( [1,2,3,4,2,4,5,2,1], 2)
2. find_all("hello好世界", "w")

所需的输出:

1. [1,4,7]
2. [6,16]

推荐答案

您可以使用 Python 的功能向后遍历列表并获取最后一个元素.然后将列表与 + 运算符放在一起.通过向后遍历列表，您可以在找到值时找到索引，而不是在从列表的开头移动到结尾时丢失它.

You can use Pythons ability to walk backwards through a list and grab the last element. Then put lists together with the + operator. By going through the list backwards you're able to find the indice when a value is found, rather than losing it when you move from the start of the list to the end.

def find_all(L, v):
    if not L:
            return []

    result = []
    if L[-1] == v:
            result = [len(L)-1]

    return find_all(L[:-1], v) + result

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