C# DFS 代码

public class Solution {
    private int e, v;
    public int CountCompleteComponents(int n, int[][] edges) {
        List<List<int>> graph = new List<List<int>>();
        for (int i = 0; i < n; i++){
            graph.Add(new List<int>());
        }

        foreach (int[] edge in edges){
            graph[edge[0]].Add(edge[1]);
            graph[edge[1]].Add(edge[0]);
        }

        int result = 0;
        bool[] visited = new bool[n];
        for (int i = 0; i < n; i++){
            if (!visited[i]){
                e = 0;
                v = 0;
                DFS(i);
                if (e == v * (v - 1)) result++;
            }
        }

        return result;

        void DFS(int x){
            v++;
            e += graph[x].Count;
            visited[x] = true;
            foreach (int y in graph[x]){
                if (!visited[y]){
                    DFS(y);
                }
            }
        }
    }
}

C# 并查集 代码

public class Solution {
    public int CountCompleteComponents(int n, int[][] edges) {
        List<List<int>> graph = new List<List<int>>();
        for (int i = 0; i < n; i++){
            graph.Add(new List<int>());
        }

        foreach (int[] edge in edges){
            graph[edge[0]].Add(edge[1]);
            graph[edge[1]].Add(edge[0]);
        }

        int[] p = new int[n];
        int[] count = new int[n];
        for (int i = 0; i < n; i++){
            p[i] = i;
            count[i] = 1;
        }

        foreach (int[] edge in edges){
            int a = edge[0], b = edge[1];
            if (find(a) == find(b)) continue;
            count[find(b)] += count[find(a)];
            p[find(a)] = find(b);
        }

        int result = 0;
        Dictionary<int, HashSet<int>> dict = new Dictionary<int, HashSet<int>>();
        for (int i = 0; i < n; i++){
            int x = find(i);
            if (!dict.ContainsKey(x)) dict.Add(x, new HashSet<int>());
            dict[x].Add(i);
        }

        foreach (var kv in dict){
            int v = count[kv.Key];
            int e = kv.Value.Sum(i => graph[i].Count);
            if (e == v * (v - 1)) result++;
        }

        return result;

        int find(int x){
            if (p[x] != x) p[x] = find(p[x]);
            return p[x];
        }
    }
}