547.省份数量.md

547.省份数量

https://leetcode-cn.com/problems/number-of-provinces/

一、并查集

连通性问题显然可以用并查集

class UnionFind:
    # 构造函数传入totalNodes为总节点数
    def __init__(self, totalNodes):
        # parents[i]表示i的根，初始i的根为其自身
        self.parents = [i for i in range(totalNodes)]
        # 连通分量数
        self.count = totalNodes
        # 树的“重量”
        self.size = [1] * totalNodes
    # 合并连通区域是通过find来操作的, 即看这两个节点是不是在一个连通区域内.
    def union(self, node1, node2):
        root1 = self.find(node1)
        root2 = self.find(node2)
        # 不连通才合并
        if root1 != root2:
          	# 小树合并到大树
            if self.size[root1] > self.size[root2]:
                self.parents[root2] = root1
                self.size[root1] += self.size[root2]
            else:
                self.parents[root1] = root2
                self.size[root2] += self.size[root1]
            # 连通分量数减一
            self.count -= 1
# 
    # 查找最终的根
    def find(self, node):
        while self.parents[node] != node:
            self.parents[node] = self.parents[self.parents[node]]
            node = self.parents[node]
        return node

    # 判断两个点是否连通
    def isConnected(self, node1, node2):
        return self.find(node1) == self.find(node2)

    # 返回连通分量个数
    def count(self):
        return self.count

class Solution:
    def findCircleNum(self, isConnected: List[List[int]]) -> int:
        n = len(isConnected)
        uf = UnionFind(n)	#一共n个节点
        for i in range(n):	#只需要遍历上三角矩阵
            for j in range(i+1, n):
                if isConnected[i][j]:
                    uf.union(i, j)
        return uf.count		#返回连通分量个数