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.
int findParent(vector<int>& parent, int k) { if (parent[k] != k) parent[k] = findParent(parent, parent[k]); return parent[k]; } vector<int> findRedundantConnection(vector<vector<int> >& edges) { vector<int> parent; for (int i = 0; i < 2001; i++) // 初始化 parent.push_back(i); int point1, point2; for (int j = 0; j < edges.size(); j++) { point1 = findParent(parent, edges[j][0]); point2 = findParent(parent, edges[j][1]); if (point1 == point2) return edges[j]; parent[point2] = point1; } return vector<int>(0, 0); }