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 undirected edge 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.
分析：UNION-FIND
1 class Solution { 2 public int[] findRedundantConnection(int[][] edges) { 3 int[] parent = new int[edges.length + 1]; 4 for (int i = 0; i < parent.length; i++) parent[i] = i; 5 6 for (int[] edge: edges){ 7 int f = edge[0], t = edge[1]; 8 if (find(parent, f) == find(parent, t)) return edge; 9 else parent[find(parent, f)] = find(parent, t); 10 } 11 12 return new int[2]; 13 } 14 15 private int find(int[] parent, int f) { 16 if (f != parent[f]) { 17 parent[f] = find(parent, parent[f]); 18 } 19 return parent[f]; 20 } 21 }
