回答此问题可获得 20 贡献值,回答如果被采纳可获得 50 分。
<p>所以我有这个清单:</p>
<pre><code>[7, 31, 31, 61, 61, 79, 29, 79, 29, 103, 37, 103, 37, 47, 47, 53, 53, 89, 5, 89, 5, 13, 13, 83, 83, 101, 53, 101, 17, 53, 11, 17, 11, 59, 17, 59, 17, 41, 41, 79, 79, 97, 3, 97, 3, 41, 19, 41, 19, 53, 53, 67, 29, 67, 29, 73, 23, 73, 23, 43, 43, 71]
</code></pre>
<p>我想知道是否有一种方法可以对它们进行排序,比如一对的最后一个元素是第二对的第一个元素,就像这样:</p>
<pre><code>[7, 31, 31, 61, 61, 79, 79, 29, 29, 103, 103, 37 ....]
[pair1] [pair2] [pair3] [pair4] [pair5] [pair6] ...
</code></pre>
<p>有办法吗?你知道吗</p>
<p>我尝试过循环,但它只是有一个非常高的时间复杂度。你知道吗</p>
<p>这是我的密码:</p>
<pre><code>
lis=[7, 31, 31, 61, 61, 79, 29, 79, 29, 103, 37, 103, 37, 47, 47, 53, 53, 89, 5, 89, 5, 13, 13, 83, 83, 101, 53, 101, 17, 53, 11, 17, 11, 59, 17, 59, 17, 41, 41, 79, 79, 97, 3, 97, 3, 41, 19, 41, 19, 53, 53, 67, 29, 67, 29, 73, 23, 73, 23, 43, 43, 71]
for i in range(0, len(lis)-4,4):
if lis[i]==lis[i+2]:
p=lis[i]
lis[i]=lis[i+1]
lis[i+1]=p
elif lis[i]==lis[i+3]:
p = lis[i]
lis[i] = lis[i + 1]
lis[i + 1] = p
p=lis[i+2]
lis[i+2]=lis[i+3]
lis[i+3]=p
elif lis[i+1]==lis[i+3]:
p=lis[i+2]
lis[i+2]=lis[i+3]
lis[i+3]=p
print(lis)
</code></pre>