给定相关数字的列表，合并相关列表以创建不相交的集合

Question

假设：

[(1,2),(3,4),(5,6),(3,7),(5,7)]

输出：

[set(1,2), set(3,4,5,6,7)]

说明：

(1,2)
(1,2), (3,4)
(1,2), (3,4), (5,6)
(1,2), (3,4,7), (5,6)
(1,2), (3,4,7,5,6)

我写了一个糟糕的算法：

Case 1: both numbers in pair are new (never seen before):
    Make a new set with these two numbers
Case 2: one of the number in pair is new, other is already a part of some set:
    Merge the new number in other's set
Case 3: both the numbers belong to some set:
    Union the second set into first. Destroy the second set.

这个算法是否有更多的pythonic（奇特）解决方案？

Answer 1

您可以使用Unionfind algorithm。首先，我们使用字典从对中创建树：

leaders = collections.defaultdict(lambda: None)

现在我们使用两个函数 - union和find - 来填充该树：

def find(x):
    l = leaders[x]
    if l is not None:
        l = find(l)
        leaders[x] = l
        return l
    return x

def union(x, y):
    lx, ly = find(x), find(y)
    if lx != ly:
        leaders[lx] = ly

迭代所有对并将它们放入树中。

for a, b in [(1,2),(3,4),(5,6),(3,7),(5,7)]:
    union(a, b)

然后看起来像这样：{1: 2, 2: None, 3: 4, 4: 7, 5: 6, 6: 7, 7: None}

最后，我们按照各自的“领导者”对这些数字进行分组，即find返回的内容：

groups = collections.defaultdict(set)
for x in leaders:
    groups[find(x)].add(x)

现在，groups.values()为[set([1, 2]), set([3, 4, 5, 6, 7])]

复杂性应该是 O（nlogn）的顺序。