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.