擅长:python、mysql、java
<p>我从两个方面低估了这个问题:</p>
<pre><code> counter = 0
for i in range(len(list_of_number)):
for j in range(i, len(list_of_number)):
if list_of_number[j] > list_of_number[i]:
counter += 1
</code></pre>
<p>在这里,在获得第一个值后,它扫描所有列表。它后面的代码将只检查号码的右邻居</p>
<pre><code> counter = 0
for i in range(len(list_of_number)-1):
if list_of_numbers[i+1] > list_of_numbers[i]:
counter +=1
</code></pre>