You are given an undirected graph. You are given an integer n which is the number of nodes in the graph and an array edges, where each edges[i] = [ui, vi] indicates that there is an undirected edge between ui and vi.
A connected trio is a set of three nodes where there is an edge between every pair of them.
The degree of a connected trio is the number of edges where one endpoint is in the trio, and the other is not.
Return the minimum degree of a connected trio in the graph, or -1 if the graph has no connected trios.
Example 1:

Input: n = 6, edges = [[1,2],[1,3],[3,2],[4,1],[5,2],[3,6]] Output: 3 Explanation: There is exactly one trio, which is [1,2,3]. The edges that form its degree are bolded in the figure above.
Example 2:

Input: n = 7, edges = [[1,3],[4,1],[4,3],[2,5],[5,6],[6,7],[7,5],[2,6]] Output: 0 Explanation: There are exactly three trios: 1) [1,4,3] with degree 0. 2) [2,5,6] with degree 2. 3) [5,6,7] with degree 2.
Constraints:
2 <= n <= 400edges[i].length == 21 <= edges.length <= n * (n-1) / 21 <= ui, vi <= nui != vi- There are no repeated edges.
思路: 用connected matrix 去build graph edge check. 然后就是暴力搜索a, b, c 。 找到之后,计算indegree,取最小的就可以了。
class Solution {
public int minTrioDegree(int n, int[][] edges) {
int[] indegree = new int[n + 1];
int[][] matrix = new int[n + 1][n + 1];
List<List<Integer>> adjlist = new ArrayList<List<Integer>>();
for(int i = 0; i < n + 1; i++) {
adjlist.add(new ArrayList<Integer>());
}
for(int[] edge: edges) {
int u = edge[0];
int v = edge[1];
indegree[u]++;
indegree[v]++;
matrix[u][v] = 1;
matrix[v][u] = 1;
adjlist.get(u).add(v);
adjlist.get(v).add(u);
}
int res = Integer.MAX_VALUE;
for(int a = 1; a <= n; a++) {
for(int i = 0; i < adjlist.get(a).size(); i++) {
for(int j = i + 1; j < adjlist.get(a).size(); j++) {
int b = adjlist.get(a).get(i);
int c = adjlist.get(a).get(j);
if(matrix[b][c] == 1) {
res = Math.min(res, indegree[a] + indegree[b] + indegree[c] - 6);
}
}
}
}
return res == Integer.MAX_VALUE ? -1: res;
}
}
