[LeetCode]684. Redundant Connection

• descrption
  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 edgesis 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]
  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.

• 解题思路
  首先有N个节点的一棵树，加上了一条边后变成了一个有环的无向图。题目让我们找到多余的一条边（删去这条边后可以得到树），如果有多条符合的边，删去最后出现的一条。
  这道题有两种思路：
  第一种，用并查集算法，下面代码就是用的这种算法。遍历每一条边，找到每个node的根节点，如果一条边的两个node的根节点是同一节点，那么这条边就是冗余的，最后找到的这条边就是要删去的边。
  第二种，其实与第一种算法类似，但是这种理解更简单，还是遍历每条边，如果边的两个node都是第一次出现或者其中一个node是第一次出现，那么这条边绝对不可能是冗余的边，不可能构成环。如果一条边的两个node在前面的边的node中都出现过，那么这条边肯定会构成环，这条边就是要找的冗余边。

• 代码如下

class Solution {
public:
    vector<int> findRedundantConnection(vector<vector<int>>& edges) {
        //root of every node, if the root is the same, then the edge is redundant
        int size = edges.size()+1;
        vector<int> root(size);
        //ans
        vector<int> ans;
        //in the begining, every node's root is itself
        for(int i = 1; i <= edges.size(); i++){
            root[i] = i;
        }
        //find the root of every node
        for(int i = 0; i < edges.size(); i++){
            Union(edges[i][0], edges[i][1], root, ans);
        }
        return ans;
    }
    void Union(int first, int second, vector<int>& root, vector<int>& ans){
        int root1 = findroot(first, root);
        int root2 = findroot(second, root);
        if(root1 != root2){
            root[root2] = root1;
        }
        else{
            ans.clear();
            ans.push_back(first);
            ans.push_back(second);
        }
    }
    int findroot(int node, vector<int>& root){
        while(node != root[node]){
            node = findroot(root[node], root);
        }
        return node;
    }

};

原题地址
如有错误请指出，谢谢！