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.

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.

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/redundant-connection
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

 

去掉无向图的一条边，让它变成一棵树。如果有多条边可以去，那么就返回数组中最后出现的边。

先找出环中的所有节点：相当于拓扑排序，每次都清除掉入度是1的节点，看最后剩下的节点，从数组中找到最后出现的。

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class Solution {
    boolean [][]graph;
    int lenght = 0;
    Stack<Pair<Integer, Integer>>stack = new Stack<>();
    public int[] findRedundantConnection(int[][] edges) {
        lenght = edges.length;
        int[]record = new int[edges.length + 1];
        for (int i = 0; i < edges.length; i++) {
            record[edges[i][0]]++;
            record[edges[i][1]]++;
        }
        graph = new boolean[edges.length + 1][edges.length + 1];
        for (int i = 0; i < edges.length; i++) {
            graph[edges[i][0]][edges[i][1]] = true;
            graph[edges[i][1]][edges[i][0]] = true;
        }

        Queue<Integer>queue = new ArrayDeque<>();
        for (int i = 1; i <= lenght; i++) {
            if (record[i] == 1) {
                queue.add(i);
                record[i]--;
            }
        }

        while (!queue.isEmpty()) {
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                int t = queue.poll();
                for (int j = 1; j <= lenght; j++) {
                    if (graph[t][j]) {
                        record[j]--;
                    }
                    if (record[j] == 1) {
                        record[j]--;
                        queue.add(j);
                    }
                }
            }
        }

        int []ans = new int[2];
        for (int i = edges.length - 1; i >= 0; i--) {
            for (int j = 1; j <= lenght; j++) {
                if (record[j] > 1) {
                    if (edges[i][0] == j || edges[i][1] == j) {
                        if (record[edges[i][0]] > 1 && record[edges[i][1]] > 1 ) {
                            ans[0] = edges[i][0];
                            ans[1] = edges[i][1];
                            return ans;
                        }
                    }
                }
            }
        }
        return null;
    }
}

​