LeetCode 684. Redundant Connection

原题链接在这里：https://leetcode.com/problems/redundant-connection/

题目：

In this problem, a tree is an undirected graph that is connected and has no cycles.

The given input is a graph that started as a tree with N nodes (with distinct values 1, 2, ..., N), with one additional edge added. The added edge has two different vertices chosen from 1 to N, and was not an edge that already existed.

The resulting graph is given as a 2D-array of edges. Each element of edges is a pair [u, v] with u < v, that represents an undirectededge connecting nodes u and v.

Return an edge that can be removed so that the resulting graph is a tree of N nodes. If there are multiple answers, return the answer that occurs last in the given 2D-array. The answer edge [u, v] should be in the same format, with u < v.

Example 1:

Input: [[1,2], [1,3], [2,3]]
Output: [2,3]
Explanation: The given undirected graph will be like this:
  1
 / 
2 - 3

Example 2:

Input: [[1,2], [2,3], [3,4], [1,4], [1,5]]
Output: [1,4]
Explanation: The given undirected graph will be like this:
5 - 1 - 2
    |   |
    4 - 3

Note:

• The size of the input 2D-array will be between 3 and 1000.
• Every integer represented in the 2D-array will be between 1 and N, where N is the size of the input array.

Update (2017-09-26):
We have overhauled the problem description + test cases and specified clearly the graph is an undirected graph. For the directed graph follow up please see Redundant Connection II). We apologize for any inconvenience caused.

题解：

For each edge, union edge[0] and edge[1]. If edge[0] and edge[1] are already in the same union. Then current edge is redundant.

Time Complexity: O(nlogn). find takes O(logn). With path compression and union by weight, amatorize O(1).

Space: O(n).

AC Java:

class Solution {
    int [] parent;
    int [] size;
    
    public int[] findRedundantConnection(int[][] edges) {
        int n = edges.length;
        parent = new int[n+1];
        size = new int[n+1];
        
        for(int i = 0; i<=n; i++){
            parent[i] = i;
            size[i] = 1;
        }
        
        for(int [] edge: edges){
            if(find(edge[0]) == find(edge[1])){
                return edge;
            }
            
            union(edge[0], edge[1]);
        }
        
        return null;
    }
    
    private int find(int i){
        if(i != parent[i]){
            parent[i] = find(parent[i]);
        }
        
        return parent[i];
    }
    
    private void union(int i, int j){
        int p = find(i);
        int q = find(j);
        
        if(size[p] > size[q]){
            parent[q] = p;
            size[p] += size[q];
        }else{
            parent[p] = q;
            size[q] += size[p];
        }
    }
}

跟上Redundant Connection II.