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number-of-connected-components-in-an-undirected-graph.py 841 Bytes
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Allen Liu 提交于 7年前 . add complexity
# Time: O(nlog*n) ~= O(n), n is the length of the positions
# Space: O(n)
class UnionFind(object):
def __init__(self, n):
self.set = range(n)
self.count = n
def find_set(self, x):
if self.set[x] != x:
self.set[x] = self.find_set(self.set[x]) # path compression.
return self.set[x]
def union_set(self, x, y):
x_root, y_root = map(self.find_set, (x, y))
if x_root != y_root:
self.set[min(x_root, y_root)] = max(x_root, y_root)
self.count -= 1
class Solution(object):
def countComponents(self, n, edges):
"""
:type n: int
:type edges: List[List[int]]
:rtype: int
"""
union_find = UnionFind(n)
for i, j in edges:
union_find.union_set(i, j)
return union_find.count
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