擅长:python、mysql、java
<pre><code>a = [0, 2, 15, 13, 13, 5, 12, 3, 1, 3, 14, 14, 17, 20, 0, 14, 2, 0,
1, 10, 5, 13, 15, 13, 19, 11, 20, 5, 4, 17, 20, 7, 4, 20, 20,
19, 3, 20, 20, 0, 14, 8, 18, 0, 18, 2, 5, 15, 12, 20, 5, 7, 12,
6, 17, 18, 15, 12, 2, 16, 6, 11, 14, 20, 8, 4, 11, 10, 15, 6,
10, 19, 1, 20, 6, 19, 2, 5, 13, 7, 5, 18, 11, 11, 3, 0, 11, 3,
14, 18, 15, 11, 2, 9, 12, 0, 18, 1, 4, 2]
</code></pre>
<p>浏览列表一次,找出所有的<em>对</em>并将它们放入一个集合中</p>
^{pr2}$
<p>然后像这样使用它:</p>
<pre><code>>>> n = 2
>>> n in z
False
>>> n = 13
>>> n in z
True
>>>
</code></pre>
<hr/>
<pre><code>zip(a, a[1:])
</code></pre>
<p>将生成<code>(a[0], a[1]), (a[1], a[2]), (a[2], a[3]), ...</code>的元组</p>
<p><code>pairwise</code><a href="https://docs.python.org/3/library/itertools.html#itertools-recipes" rel="nofollow noreferrer">Itertools Recipe</a>将执行相同的操作。在</p>