擅长:python、mysql、java
<p>在这种情况下,我们可以使用自定义计数函数,从感兴趣的字符串中增加一个子字符串,直到找到匹配项,然后重置它:</p>
<pre class="lang-py prettyprint-override"><code>def count(string, *substrings):
acc = '' # accumulator
matches = 0
for char in string:
acc += char
for substring in substrings:
if substring in acc:
matches += 1
acc = ''
break
return matches
</code></pre>
<p>我们这样称呼它:</p>
<pre class="lang-py prettyprint-override"><code>count('abbabba', 'ab', 'ba') # 3
</code></pre>