[LeetCode]Number of Connected Components in an Undirected Graph

最新推荐文章于 2025-08-22 01:16:35 发布

转载最新推荐文章于 2025-08-22 01:16:35 发布 · 90 阅读

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原文链接：https://segmentfault.com/a/1190000004224298

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#数据结构与算法

本文介绍了一种使用并查集算法解决无向图中连通分量计数问题的方法。通过遍历图中的边进行节点合并，最终统计独立根节点数量来确定连通分量的数目。

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Number of Connected Components in an Undirected Graph

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.

分析

典型且很基础的union find题。用一个数组记录各个数字的父节点，然后遍历图，对edge中两个端点做union。最后扫一遍数组，找到根节点个数即可。

time:

time: O(m*h), space: O(n), m表示edge的数量。

代码

public class Solution {
    public int countComponents(int n, int[][] edges) {
        int[] id = new int[n];
        
        // 初始化
        for (int i = 0; i < n; i++) {
            id[i] = i;
        }
        
        // union
        for (int[] edge : edges) {              
            int i = root(id, edge[0]);
            int j = root(id, edge[1]);
            id[i] = j;
        }
        
        // 统计根节点个数
        int count = 0;
        for (int i = 0; i < n; i++) {
            if (id[i] == i)
                count++;
        }
        return count;
    }
    
    // 找根节点
    public int root(int[] id, int i) {
        while (i != id[i]) {
            id[i] = id[id[i]];
            i = id[i];
        }
        return i;
    }
}