1.题目要求
link:LeetCode684
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.
2.分析:
简单来说就是从输入的二维数组构成的图中找出最后那条使这个图有环的那条边。
注意：在题目所给出的提示中我们得知，输入二维数组中的最大值是输入数组的长度。
这时就会考虑是否可以用一个unordered_set数据结构来解决问题，通过判断连接的两个点是否都在集合中来判断是否这条边构成环。事实证明这种方法是不可以的，因为如果图的构成顺序不是在同一个集合中就会出错。
那可否用查并集来做呢？只要在一个集合中就会构成环，同时可以存在多个集合，并且当两个集合有边连接时还可以合并两个集合。通过分析，应该是正确的，那就试验一下吧！
3.结果
通过LeetCode验证测试通过！