擅长:python、mysql、java
<p>下面是一个简单的暴力解决方案:</p>
<pre><code>from collections import Counter
s = " HIYourName=this is not true HIYourName=Have a good day HIYourName=nope HIYourName=Bye!"
for n in range(1, len(s)):
substr_counter = Counter(s[i: i+n] for i in range(len(s) - n))
phrase, count = substr_counter.most_common(1)[0]
if count == 1: # early out for trivial cases
break
print 'Size: %3d: Occurrences: %3d Phrase: %r' % (n, count, phrase)
</code></pre>
<p>示例字符串的输出为:</p>
^{pr2}$