Leetcode/G家twitter -- 323. Number of Connected Components in an Undirected Graph (Union Find)

323. Number of Connected Components in an Undirected Graph(Union Find)

• Difficulty: Medium

https://leetcode.com/problems/number-of-connected-components-in-an-undirected-graph/

iven 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.

参考：https://discuss.leetcode.com/topic/32752/easiest-2ms-java-solution

This is 1D version of Number of Islands II. For more explanations, check out this 2D Solution.

1. n points = n islands = n trees = n roots.
2. With each edge added, check which island is e[0] or e[1] belonging to.
3. If e[0] and e[1] are in same islands, do nothing.
4. Otherwise, union two islands, and reduce islands count by 1.
5. Bonus: path compression can reduce time by 50%.

Hope it helps!

Complexity: The complexity for M quick union + path compression on N objects should be N + MlgN, I think in this problem, M = 2E, N = V, so O(V + 2ElgV)

public class Solution {
    public int countComponents(int n, int[][] edges) {
         // initialize n isolated islands
        int[] roots = new int[n];
        for(int i = 0; i < n; i++) roots[i] = i; 
        int count =n;
        // perform union find
        for (int i = 0; i < edges.length; i++) {
            int root1 = findSet(roots,edges[i][0]);//when union, always update the top parent(recur)
            int root2 = findSet(roots,edges[i][1]);
            //if in one set means already reduce count by adding edge before, new edge not impact
            if(root1!=root2){
                roots[root1] = root2;//in 2 sets,union(add a edge)
                count--;
            }
        }
        return count;
    }
    public int findSet(int[] roots,int i){
        if(roots[i]==i)return i;//is top parent
        roots[i] = findSet(roots,roots[i]);//path compression(optional)
        return roots[i];
    }
}

Optional path compression: (Iteration has little difference from recursion)

public int findSet(int[] roots,int i){
        while(roots[i] != i) {
            roots[i] = roots[roots[i]];  // optional: path compression
            i = roots[i];
        }
        return i;
    }