给定相关数字列表，合并相关列表以创建不相交的集合

Question

鉴于：

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

输出：

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

解释：

(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.

这个算法有没有更Pythonic（奇特）的解决方案？

Answer 1

为此，您可以使用Unionfind 算法。首先，我们使用字典从对中创建一棵树：

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

只需迭代所有对并将它们放入树中即可。

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

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

最后，我们按各自的“领导者”对数字进行分组，即返回的内容find：

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

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

复杂度应该约为O(nlogn)。