如何使用递归查找列表中偶数的总和?
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.
答
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])