题目
undirected
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:
Example 2:
Input: [[1,2], [2,3], [3,4], [1,4], [1,5]] Output: [1,4] Explanation:
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.
思路
这道题目是道典型的union-find问题，这个动态连通性问题，大家可以参考我以前分享的一篇解题报告，基本好多问题都是从那个问题衍生而来。
代码
class Solution {
public:
vector<int> findRedundantConnection(vector<vector<int>>& edges) {
int M= edges.size();
vector<int> id(M+1);
vector<int> res;
for(int i=0;i<=M;i++)
id[i] = i;
for(auto edge:edges){
int p = edge[0];
int q = edge[1];
while(p!=id[p])
p = id[p];
while(q!=id[q])
q = id[q];
if(p==q)
res = edge;
else
id[q] = p;
}
return res;
}
};
