[LeetCode] 684. Redundant Connection

lncwwn 发布于2019-08-16 13:55 / 291人阅读

Problem

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.

Solution

class Solution {
    public int[] findRedundantConnection(int[][] edges) {
        UnionFind uf = new UnionFind(edges.length+1);
        for (int[] edge: edges) {
            int a = edge[0], b = edge[1];
            if (uf.find(a) == uf.find(b)) return edge;
            else uf.union(a, b);
        }
        return new int[0];
    }
}
class UnionFind {
    int[] parents;
    public UnionFind(int n) {
        parents = new int[n];
        for (int i = 0; i < n; i++) parents[i] = i;
    }
    public int find(int a) {
        if (parents[a] != a) return find(parents[a]);
        else return a;
    }
    public void union(int a, int b) {
        int pA = find(a), pB = find(b);
        if (pA != pB) parents[pB] = pA;
    }
}

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