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.
分析:UNION-FIND
class Solution {
public int[] findRedundantConnection(int[][] edges) {
int[] parent = new int[edges.length + ];
for (int i = ; i < parent.length; i++) parent[i] = i; for (int[] edge: edges){
int f = edge[], t = edge[];
if (find(parent, f) == find(parent, t)) return edge;
else parent[find(parent, f)] = find(parent, t);
} return new int[];
} private int find(int[] parent, int f) {
if (f != parent[f]) {
parent[f] = find(parent, parent[f]);
}
return parent[f];
}
}
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