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.
class Solution { public: vector<int> findRedundantConnection(vector<vector<int>>& edges) { // union find 方法 vector<int> root(2001, -1); for (auto& edge : edges) { int x = findIfConnected(root, edge[0]), y = findIfConnected(root, edge[1]); if (x == y) return edge; root[x] = y; } return {}; } int findIfConnected(vector<int>& root, int i) { while (root[i] != -1) { i = root[i]; } return i; } };
参考： https://github.com/grandyang/leetcode/issues/684
转载于:https://www.cnblogs.com/simplepaul/p/11336029.html
