684. Redundant Connection

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.

分析：

字面意思是给一个图由n条边构成，去掉某条边后成为了树（无向无环连通图）

题意是给一个图，由一条边一条边构成，当加入某条边后形成了环，我们现在要找到这条边。

采用查并集（并查集（union find））来做，本质是一个数据结构（class)，里面有int[ ] parent, int[ ] rank, 分别对应的是该数字的parent和rank，rank有点像height（size），所以可以union by rank OR union by size.

　　查int find(x)：找出x的parent并返回，具体实现是直到parent是它自己为止（因为初始化每个数parent都是他自己）

　　并 boolean union(int x, int y)：尝试合并x和y，如果能合并说明这条边的加入没有构成环，如果不能，说明他们拥有相同的parent，那很明显加入这条边会成环。

具体实现是比较两个的parent，接着要compress一下，把rank低的合并到rank高的,记得更新rank

初始化时多一位，因为数字是1-N

class Solution {
    public int[] findRedundantConnection(int[][] edges) {
            DS ds = new DS(edges.length+1);

            for (int[] edge : edges) {
                if (!ds.union(edge[0], edge[1])) return edge;
            }
            return new int[]{};
        }

        static class DS {

            private int[] parent;
            private int[] rank;

            public DS(int n) {
                parent = new int[n];
                rank = new int[n];
                
                for(int i = 0; i < n; i++) parent[i] = i;
            }

            public int find(int x) {
                if(parent[x] != x){
                    parent[x] = find(parent[x]);
                }
                return parent[x];
            }

            // Return false if x, y are connected.
            public boolean union(int x, int y) {
                int rootX = find(x);
                int rootY = find(y);
                if (rootX == rootY) return false;

                // Make root of smaller rank point to root of larger rank.
  
                if(rank[rootX] > rank[rootY]){
                    parent[rootY] = rootX;
                    rank[rootX]+=rank[rootY];
                }
                else{
                    parent[rootX] = rootY;
                    rank[rootY]+=rank[rootX];
                }
                return true;
            }
        }
}

https://leetcode.com/problems/redundant-connection/discuss/123819/Union-Find-with-Explanations-(Java-Python)

union find看下https://www.geeksforgeeks.org/union-find-algorithm-set-2-union-by-rank/或者花花酱