如何使用递归查找列表中偶数的总和? [英] How to find sum of even numbers in a list using recursion?

问题描述

def sum_evens_2d(xss):
    i = 0
    counter = 0
    while (i < len(xss)):
        if(xss[i]%2 == 0):
            counter += xss[i]   
            i= i+1
        else:
            i = i+1
    return(counter)

我试图在 xss 列表中找到偶数的总和.我的限制是我不能使用 sum()，而只能使用递归.

I am trying to find the sum of the evens in the list xss. My restrictions are that I can not use sum(), but recursion only.

推荐答案

刚刚测试了这个，应该可以:

Just tested this one out, it should work:

def even_sum(a):
    if not a:
        return 0
    n = 0
    if a[n] % 2 == 0:
        return even_sum(a[1:]) + a[n]
    else:
        return even_sum(a[1:])

# will output 154
print even_sum([1, 2, 3, 4, 5, 6, 7, 8, 23, 55, 45, 66, 68])

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