擅长:python、mysql、java
<p>我会这样解决:</p>
<pre><code>def count_zero_one_two(s):
num = 0
for i in range(len(s)):
for j in range(1, len(s)/3 + 1):
if all(s[i:i+3*j].count(n) == j for n in '012'):
num += 1
return num
</code></pre>
<p><a href="https://docs.python.org/3/library/functions.html#all" rel="nofollow noreferrer">^{<cd1>}</a>用于检查所有3个字符(对于每个迭代)是否都在“012”中。在</p>
<p>内部<code>for</code>循环用于按长度3、6、9等顺序计算0、1和2的数目</p>
<p><strong>输出:</strong></p>
^{pr2}$