LeetCode|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.

思路：树有n个节点，n-1条边。每增加一条边，应该把一个新的节点连接到树上。如果一条边连接的两个节点都属于同一个集合，也就是都连在同一颗树上了，那就是冗余的边。

class Solution {
public:
    vector<int> findRedundantConnection(vector<vector<int> >& edges) {
        // tree: n node, n-1 edges
        vector<int> res;
        int n = edges.size();
        if(n == 0) return res;
        parent = vector<int>(2000);  // 写n会报错
        rank = vector<int>(2000, 0);
        // make set
        for(int i = 0; i < n; ++i) parent[i] = i;
        for(int i = 0; i < n; ++i){
            if(find(edges[i][0]) == find(edges[i][1])) {
                return edges[i];
                //res.push_back(edges[i][0]);
                //res.push_back(edges[i][1]);
                //return res;
            }
            union_set(edges[i][0], edges[i][1]);
        } return res;
    }
private:
    vector<int> parent, rank;
    int find(int x){
        if(x != parent[x]) parent[x] = find(parent[x]);
        return parent[x];
    }
    void union_set(int x, int y){
        if((x = find(x)) == (y = find(y))) return;
        if(rank[x] > rank[y]) parent[y] = x;
        else {
            parent[x] = y;
            if(rank[x] == rank[y]) ++rank[y];
        }
    }
};