684. Redundant Connection

Leetcode Tree Union Find Graph

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¶

这道题目考查图论的基本操作-检测无向图的是否有环。类似于LeetCode261. Graph Valid Tree，换汤不换药。具体方法是：使用并查集存放连通域，将每一条边的两个节点执行并(union)操作，如果存在环，那么这两个顶点一定已经在同一连通域中，返回这条边；否则不存在环。

class Solution {
    public int[] findRedundantConnection(int[][] edges) {
        int n = edges.length;
        int[] id = new int[n + 1], size = new int[n + 1];
        for (int v = 1; v <= n; v++) id[v] = v;
        int v, w, i, j;
        for (int[] edge : edges) {
            v = edge[0]; w = edge[1];
            i = find(id, v); j = find(id, w);
            if (i == j) return new int[]{v, w};
            union(id, size, v, w);
        }
        return new int[]{-1, -1};
    }


    private void union(int[] id, int[] size, int v, int w) {
        int i = find(id, v);
        int j = find(id, w);
        if (i == j) return;
        if (size[i] > size[j]) {
            id[j] = i;
            size[i] += size[j];
        } else {
            id[i] = j;
            size[j] += size[i];
        }
    }

    private int find(int[] id, int i) {
        while (i != id[i]) {
            id[i] = id[id[i]];
            i = id[i];
        }
        return i;
    }

}