684. 冗余连接

684. 冗余连接

rank优化的并查集，顺序遍历edges，如果边在同一连通分量，说明这条边是冗余的。

时间复杂度：O（n）

空间复杂度：O（n）

class Solution {
    public int[] findRedundantConnection(int[][] edges) {
        int n = edges.length;
        Disjoint_Set disjoint_Set = new Disjoint_Set(n+1);
        for(int[] path : edges) {
        	int from = path[0];
        	int to = path[1];
        	if(!disjoint_Set.isSame(from, to)) {
        		disjoint_Set.union(from, to);
        	}else {
        		return path;
        	}
        }
        return new int[] {};
    }
    class Disjoint_Set {
    	int[] parents;
    	int[] rank;
    	public Disjoint_Set(int n) {
    		parents = new int[n];
    		rank = new int[n];
    		for (int i = 0; i < n; i++) {
    			parents[i] = i;
    			rank[i] = 1;	//初始每个集合大小为1
    		}
    	}

    	public int find(int x) {
    		if(parents[x] == x)
    			return x;
    		return find(parents[x]);
    	}

    	public boolean isSame(int x, int y) {
    		return find(x) == find(y);
    	}

    	public void union(int x, int y) {
    		int p1 = find(x);
    		int p2 = find(y);
    		if(p1 == p2)	//所属同一集合,直接返回
    			return ;
    		if(rank[p1] < rank[p2]) {	//树矮的连接到树高的，rank不变
    			parents[p1] = p2;
    		}else if (rank[p1] > rank[p2]){	
    			parents[p2] = p1;
    		}else {	//只有树高相同时,连接的时候,连接的根节点rank+1
    			parents[p2] = p1;
    			rank[p1]++ ;
    		}
    		return;
    	}
    }

}