lc 684 Redundant Connection
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.
Union Find Accepted
Union Find的关键思想是使各结点依次连结在一起，如果有从1到2的边，则令uni[1] = 2，如果有从2到3的边，则令uni[2] = 3，这样如果此时新加一条从1到3的边，那么从1开始寻找，uni1为2，uni[2]为3，即1已经有了一条从1到3的边，再加会导致形成环路，所以这就是我们要的答案。
class Solution {
public:
vector<int> findRedundantConnection(vector<vector<int>>& edges) {
vector<int> uni(2001, -1);
for (auto edge : edges) {
int head = edge[0], tail = edge[1];
int x = find(uni, head), y = find(uni, tail);
if (x == y) return edge;
uni[x] = y;
}
return {};
}
int find(vector<int>& uni, int num) {
while(uni[num] != -1) {
num = uni[num];
}
return num;
}
};