边工作边刷题：70天一遍leetcode: day 94
Number of Connected Components in an Undirected Graph
要点：
union-find如何记？
- data structure就一个array/list，每个元素初始化为val=index，表示每个元素是独立的。之后还保持的元素一定是union后的根。
- union的过程先find（O(lgn)），然后选rank小的合并。
- 这里可以简化，不记录rank，任意合并，code可以简化不少
错误点：
- find：recursion的arg是parents[x]，不是x（否则成死循环了）
- xrange doesn’t support item assignment, use range instead
# Given n nodes labeled from 0 to n - 1 and a list of undirected edges (each edge is a pair of nodes), write a function to find the number of connected components in an undirected graph.
# Example 1:
# 0 3
# | |
# 1 --- 2 4
# Given n = 5 and edges = [[0, 1], [1, 2], [3, 4]], return 2.
# Example 2:
# 0 4
# | |
# 1 --- 2 --- 3
# Given n = 5 and edges = [[0, 1], [1, 2], [2, 3], [3, 4]], return 1.
# Note:
# You can assume that no duplicate edges will appear in edges. Since all edges are undirected, [0, 1] is the same as [1, 0] and thus will not appear together in edges.
class Solution(object):
def countComponents(self, n, edges):
"""
:type n: int
:type edges: List[List[int]]
:rtype: int
"""
p = range(n)
def find(x):
if p[x]!=x:
p[x]=find(p[x])
return p[x]
for s,d in edges:
ss, dd = find(s),find(d)
p[ss] = dd
n-=ss!=dd
return n
sol = Solution()
assert sol.countComponents(5, [[0,1],[1,2],[3,4]])==2, "should be 2"
assert sol.countComponents(5, [[0,1],[1,2],[0,2],[3,4]])==2, "should be 2"
