擅长:python、mysql、java
<p>这些问题更多地围绕着数学和算法而不是pythonisms。我在下面提出的解决方案在<code>O(n**2)</code>中具有复杂性。你知道吗</p>
<p>其思想是反转函数(x,y)=>;x*x+y*y,其中搜索空间是原始列表与其自身的叉积。然后,使用Python集操作符,计算应用程序图像和可接受的正方形之间的交集。最后,使用反向应用程序重建三胞胎。你知道吗</p>
<pre><code>from collections import defaultdict
original_list = [8, 5, 73, 3, 34, 4, 23, 73]
uniq = sorted(set(original_list))
antecedents = defaultdict(lambda: []) # Reverse mapping
for i, left in enumerate(uniq):
for right in uniq[i+1:]:
key = left * left + right * right
antecedents[key].append((left, right))
# The keys of antecedents are sum of squares
uniq_squares = set([ x * x for x in uniq ])
common_keys = uniq_squares & antecedents.keys()
for key in common_keys:
sqrt = int(0.5 + key**0.5)
key_antecedents = antecedents[key]
for (left, right) in key_antecedents:
print("Found triplet:", (left, right, sqrt))
</code></pre>