擅长:python、mysql、java
<pre><code>def findprevpalindrom(numb):
s = str(numb - 1)
l = len(s)
if s[:l//2][::-1] > s[(l+1)//2:]:
top = str(int(s[:(l+1)//2])-1)
else:
top = s[:(l+1)//2]
if len(top) < l/2:
return int(top + '9' + top[:l//2][::-1]) #tricky part
else:
return int(top + top[:l//2][::-1]) #the generic return
</code></pre>
<p>上述方法应能提供所需的输出</p>
<p>findprevpalindrom(100)->;九十九</p>
<p>findprevpalindrom(99)——>;88</p>
<p>findprevpalindrom(11)——>;九,</p>