LeetCode 323. Number of Connected Components in an Undirected Graph - 彻底掌握并查集（Union Find）系列题16

最新推荐文章于 2025-04-11 12:16:50 发布

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最新推荐文章于 2025-04-11 12:16:50 发布

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分类专栏： Leetcode刷题笔记并查集Union Find 文章标签： leetcode 算法数据结构 python

本文链接：https://blog.youkuaiyun.com/hgq522/article/details/121690399

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该博客介绍了如何运用并查集（UnionFind）算法解决图的连通组件计数问题。通过给出的示例代码，展示了如何在Python中实现并查集数据结构，并应用在给定的图数据上，以计算连通组件的数量。题目与LeetCode的200题NumberofIslands的解法相似，是学习并查集的入门实例。

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You have a graph of n nodes. You are given an integer n and an array edges where edges[i] = [ai, bi] indicates that there is an edge between ai and bi in the graph.

Return the number of connected components in the graph.

Example 1:

Input: n = 5, edges = [[0,1],[1,2],[3,4]]
Output: 2

Example 2:

Input: n = 5, edges = [[0,1],[1,2],[2,3],[3,4]]
Output: 1

Constraints:

1 <= n <= 2000
1 <= edges.length <= 5000
edges[i].length == 2
0 <= ai <= bi < n
ai != bi
There are no repeated edges.

一道用于并查集Union Find的入门上手题，解法跟Leetcode200. Number of Islands几乎一模一样，套用Union Find模板几乎不要改动。对每条边的两个顶点进行合并，最后集合个数就是答案。

class UnionFind:
    def __init__(self, n):
        self.parent = list(range(0, n))
        self.size = [1] * n
        self.cnt = n
    
    def find(self, x):
        if self.parent[x] != x:
            self.parent[x] = self.find(self.parent[x])
        return self.parent[x]
    
    def merge(self, x, y):
        px, py = self.find(x), self.find(y)
        if px == py:
            return
        
        if self.size[px] > self.size[py] :
            self.parent[py] = px
            self.size[px] += self.size[py]
        else:
            self.parent[px] = py
            self.size[py] += self.size[px]
                
        self.cnt -= 1
    
class Solution:
    def countComponents(self, n: int, edges: List[List[int]]) -> int:
        uf = UnionFind(n)
        
        for e in edges :
            uf.merge(e[0], e[1])
        
        return uf.cnt

可以在刷LeetCode 684. Redundant Connection之前先刷一遍这道题。