擅长:python、mysql、java
<p>我会在列表上循环,没有第一个和最后一个元素,并将每个元素与它前面的元素和后面的元素进行比较</p>
<pre><code>turn = None # i.e., no turning point
for i in range(1, len(a) - 1):
if (a[i+1] > a[i] and a[i] < a[i-1]) or (a[i+1] < a[i] and a[i] > a[i-1]):
turn = i
</code></pre>