684. Redundant Connection

We are given a "tree" in the form of a 2D-array, with distinct values for each node.

In the given 2D-array, each element pair [u, v] represents that v is a child of u in the tree.

We can remove exactly one redundant pair in this "tree" to make the result a tree.

You need to find and output such a pair. If there are multiple answers for this question, output the one appearing last in the 2D-array. There is always at least one answer.

Example 1:

Input: [[1,2], [1,3], [2,3]]
Output: [2,3]
Explanation: Original tree will be like this:
  1
 / \
2 - 3

Example 2:

Input: [[1,2], [1,3], [3,1]]
Output: [3,1]
Explanation: Original tree will be like this:
  1
 / \\
2   3

Note:

• The size of the input 2D-array will be between 1 and 1000.
• Every integer represented in the 2D-array will be between 1 and 2000.
  

  思路：类似并查集之类的东西

  package l684;
  class Solution {
      public int[] findRedundantConnection(int[][] edges) {
          int[]a = new int[2000+1];
          int cnt = 1;
          
          for(int[] edge : edges) {
          	int i=edge[0], j=edge[1];
          	if(a[i]==0 && a[j]==0) {
          		a[i] = cnt;
          		a[j] = cnt;
          		cnt++;
          	}else if(a[i] != 0 && a[j]!=0) {
          		if(a[i]==a[j]) return edge;
          		else {
          			int tmp = a[j];
          			for(int k=0; k<a.length; k++) {
          				if(a[k] == tmp)
          					a[k] = a[i];
          			}
          		}
          	} else {
          		if(a[i] == 0)	a[i] = a[j];
          		else	a[j] = a[i];
          	}
          }
          
          return null;
      }
  }