擅长:python、mysql、java
<p>你的方法没有时间效率,是二次时间。如果你考虑到线性空间,你可以在线性时间内完成。创建<code>list2</code>中元素到其索引的<em>映射</em>:</p>
<pre><code># assuming unique elements, you can modify to use a lists of indices as values to handle
l1 = [1,2,3,4,5]
l2 = [3,4,1,2,5]
index_map = {val:i for i, val in enumerate(l2)}
for val in l1:
if val in l2:
print(f"Matched {val} at position {index_map[val]}")
</code></pre>