擅长:python、mysql、java
<p>两个可能的改进:第一,如果总是只有一个参与者没有完成</p>
<pre><code>def solution(participant, completion):
for p in participant:
if p not in completion:
print( p )
return p
</code></pre>
<p>第二,如果n名参与者未完成:</p>
<pre><code>def solution(participant, completion):
answer = []
for p in participant:
if p not in completion:
print( p )
answer.add(append)
#print( answer ) # if needed
return answer
</code></pre>
<p>如果使用集合,则可以简单地跟踪它们</p>
<pre><code>def short(participant, completion):
return set(participant) - set(completion)
</code></pre>
<p>见:</p>
<ul>
<li><a href="https://docs.python.org/2/library/sets.html" rel="nofollow noreferrer">https://docs.python.org/2/library/sets.html</a></li>
<li><a href="https://www.w3schools.com/python/python_sets.asp" rel="nofollow noreferrer">https://www.w3schools.com/python/python_sets.asp</a></li>
</ul>
<p>一个小提示:
也是你的
<code>answer = print ('"' + str(*participant) + '"')</code>只适用于Python3,Python2会引发一个语法错误,因为print不会返回任何内容(Python3中没有的内容)</p>