擅长:python、mysql、java
<p>一种方法是<code>sorted</code>和<a href="https://docs.python.org/3/library/itertools.html#itertools.groupby" rel="nofollow noreferrer">^{<cd2>}</a>:</p>
<pre><code>[list(v) for _ ,v in groupby(sorted(tokens, key=len), key=len)]
[[[46565], [44460]],
[[73, 2062],
[1616, 338],
[9424, 24899],
[1820, 11268],
[43533, 5356],
[9930, 1053]],
[[260, 259, 1151]],
[[83, 31840, 292, 3826]]]
</code></pre>