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
dfs的思路是：无论递归栈多深，都只计算为1
class Solution { public int countComponents(int n, int[][] edges) { List<Integer>[] graph = new List[n]; for (int i = 0; i < n; i++) graph[i] = new ArrayList<>(); for (int[] e : edges) { graph[e[0]].add(e[1]); graph[e[1]].add(e[0]); } int components = 0; boolean[] visited = new boolean[n]; for (int v = 0; v < n; v++) components += dfs(v, graph, visited); return components; } int dfs(int u, List<Integer>[] graph, boolean[] visited) { if (visited[u]) return 0; visited[u] = true; for (int v : graph[u]) dfs(v, graph, visited); return 1; } }
323. Number of Connected Components in an Undirected Graph 连通图的数量
原文：https://www.cnblogs.com/immiao0319/p/15195894.html