擅长:python、mysql、java
<p>用适当的索引继续调用<code>a.index</code>上的<code>b</code>。如果索引是子集的开始,则处于相同的运行中。否则,开始新的运行:</p>
<pre><code>def longest_run(string, pattern):
longest = 0
current = 0
start = 0
while True:
try:
ind = string.index(pattern, start)
if ind == start:
current += 1
else:
if current > longest:
longest = current
current = 1
start += len(pattern)
except ValueError:
return longest
</code></pre>