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.
經典union find題型。
class Solution {
public int[] findRedundantConnection(int[][] edges) {
if (edges == null || edges.length == 0 || edges[0].length == 0) return new int[]{};
int n = edges.length;
int m = edges[0].length;
int[] union = new int[n+1];
for (int i = 1; i <= n; i++) union[i] = i;
for (int i = 0; i < n; i++) {
int leftRoot = find(edges[i][0], union);
int rightRoot = find(edges[i][1], union);
if (leftRoot == rightRoot) return new int[]{edges[i][0], edges[i][1]};
union[leftRoot] = rightRoot;
}
return new int[]{};
}
public int find(int n, int[] union) {
while (n != union[n]) n = union[n];
return n;
}
}