擅长:python、mysql、java
<p>我的解决办法比你的简单。而不是把字母“b”“o”“b”分开。切片字符串<strong>s</strong></p>
<pre><code>countbob=0
bob='bob'
for i in range(len(s)):
bob2=s[i:i+3]
if bob2 == bob:
countbob +=1
print ("Number of times bob occurs is: " + str(countbob))
</code></pre>