擅长:python、mysql、java
<p>这更像是一种函数方法,其中答案是通过递归函数传递的</p>
<pre><code>lst1 = [1, 2, 2013, 23, 9, 2011, 10, 11, 2000]
lst2 = []
lst3 = [1,2, 2015]
lst4 = [1,2]
lst5 = [1]
lst6 = [1,2,2013, 23, 9]
def groupToDate(lst, acc):
if len(lst) < 3:
return acc
else:
# take first elements in list
day = lst[0]
month = lst[1]
year = lst[2]
acc.append(str(day) + '/' + str(month) + '/' + str(year))
return groupToDate(lst[3:len(lst)], acc)
print(groupToDate(lst1, []))
print(groupToDate(lst2, []))
print(groupToDate(lst3, []))
print(groupToDate(lst4, []))
print(groupToDate(lst5, []))
print(groupToDate(lst6, []))
</code></pre>
<p>如果您不想使用列表理解或groupby,这也是解决此类问题的基本方法</p>