Redundant Connection(leetcode)

Redundant Connection

---

题目

leetcode题目
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

---

解决

class Solution {public:    int parent[10000] = {};    int findParent(int node) {        if (parent[node] == node) {            return node;        } else {            int result = findParent(parent[node]);            parent[node] = result;            return result;        }    }    vector<int> findRedundantConnection(vector<vector<int>>& edges) {        vector<int> result;        for (int i = 0; i < 10000; i++) {            parent[i] = i;        }        for (int i = 0; i < edges.size(); i++) {            int left = findParent(edges[i][0]);            int right = findParent(edges[i][1]);            if (left != right) {                parent[right] = left;            } else {                result.clear();                result = edges[i];            }        }        return result;    }};