684. 冗余连接 

 1 class UnionFind 
 2 {
 3 public:
 4     vector<int> parent; // 存储若干棵树
 5     int count;
 6     UnionFind(int n) 
 7     {
 8         count = n;
 9         for (int i = 0; i < n; i++)  parent.push_back(i);
10     }
11     /* 返回节点 x 的根节点 */
12     int Find(int x) 
13     {
14         // 进行路径压缩
15         if (parent[x] != x) parent[x] = Find(parent[x]);
16         return parent[x];
17     }
18     void Union(int p, int q) 
19     {
20         int rootP = Find(p);
21         int rootQ = Find(q);
22         if (rootP == rootQ) return;
23 
24         parent[rootQ] = rootP;
25         count--;
26     }
27 };
28 
29 class Solution 
30 {
31 public:
32     vector<int> findRedundantConnection(vector<vector<int>>& edges) 
33     {
34         if(edges.empty() || edges[0].empty()) return {};
35         int n = edges.size();
36         UnionFind uf(n + 1);
37         for(auto e : edges)
38         {
39             int a = e[0],b = e[1];
40             if(uf.Find(a) == uf.Find(b)) return {a,b};
41             uf.Union(a,b);
42         }
43         return {};
44     }
45 };

 

posted @ 2020-04-09 21:25  Jinxiaobo0509  阅读(178)  评论(0)    收藏  举报