擅长:python、mysql、java
<p>以下是您可以做的:</p>
<pre><code>from collections import *
def compress(s):
res= Counter(s)
l1 = sorted(res.most_common(), key=lambda t: t[0])
l2 = sorted(l1, key=lambda t: t[1], reverse=True)
print('\n'.join([f"{t[0]} {t[1]}" for t in l2]))
compress("hhgoogle")
</code></pre>
<p>输出:</p>
<pre><code>g 2
h 2
o 2
e 1
l 1
</code></pre>