LeetCode 323. Number of Connected Components in an Undirected Graph

原题链接在这里：https://leetcode.com/problems/number-of-connected-components-in-an-undirected-graph/

题目：

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.

Note:
You can assume that no duplicate edges will appear in edges. Since all edges are undirected, [0, 1] is the same as [1, 0] and thus will not appear together in edges.

题解：

使用一维UnionFind.

Time Complexity: O(elogn). e是edges数目. Find, O(logn). Union, O(1).

Space: O(n).

AC Java:

 class Solution {
     public int countComponents(int n, int[][] edges) {
         if(n <= 0){
             return 0;
         }

         UnionFind uf = new UnionFind(n);
         for(int [] edge : edges){
             if(!uf.find(edge[0], edge[1])){
                 uf.union(edge[0], edge[1]);
             }
         }

         return uf.size();
     }
 }

 class UnionFind{
     private int count;
     private int [] parent;
     private int [] size;

     public UnionFind(int n){
         this.count = n;
         parent = new int[n];
         size = new int[n];
         for(int i = 0; i<n; i++){
             parent[i] = i;
             size[i] = 1;
         }
     }

     public boolean find(int i, int j){
         return root(i) == root(j);
     }

     private int root(int i){
         while(i != parent[i]){
             parent[i] = parent[parent[i]];
             i = parent[i];
         }

         return parent[i];
     }

     public void union(int i, int j){
         int rootI = root(i);
         int rootJ = root(j);
         if(size[rootI] > size[rootJ]){
             parent[rootJ] = rootI;
             size[rootI] += size[j];
         }else{
             parent[rootI] = rootJ;
             size[rootJ] += size[rootI];
         }

         this.count--;
     }

      public int size(){
         return this.count;
     }
 }

也可以使用BFS, DFS.

Time Complexity: O(n+e), 建graph用O(n+e), BFS, DFS 用 O(n+e). Space: O(n + e).

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

         for(int [] edge : edges){
             graph.get(edge[0]).add(edge[1]);
             graph.get(edge[1]).add(edge[0]);
         }

         HashSet<Integer> visited = new HashSet<Integer>();
         int count = 0;
         for(int i = 0; i<n; i++){
             if(!visited.contains(i)){
                 // bfs(graph, i, visited);
                 dfs(graph, i, visited);
                 count++;
             }
         }
         return count;
     }

     public void bfs(List<List<Integer>> graph, int i, HashSet<Integer> visited){
         LinkedList<Integer> que = new LinkedList<Integer>();
         visited.add(i);
         que.add(i);
         while(!que.isEmpty()){
             int cur = que.poll();
             List<Integer> neighbours = graph.get(cur);
             for(int neighbour : neighbours){
                 if(!visited.contains(neighbour)){
                     que.add(neighbour);
                     visited.add(neighbour);
                 }
             }
         }
     }

     public void dfs(List<List<Integer>> graph, int i, HashSet<Integer> visited){
         visited.add(i);
         for(int neighbour : graph.get(i)){
             if(!visited.contains(neighbour)){
                 dfs(graph, neighbour, visited);
             }
         }
     }
 }

跟上Find the Weak Connected Component in the Directed Graph, Number of Islands II.