1319. 连通网络的操作次数

解法一：并查集

先判断线和结点数量的大小，线的数量小于结点数量-1则必然无法连通。
然后就是一个并查集问题，有 K 个独立的连通分支，就需要 K-1 次操作。

import java.util.*;
class Solution {
    private int[] father;
    public int makeConnected(int n, int[][] connections) {
        int lineNum = connections.length;
        if (lineNum < n - 1) {
            return -1;
        }
        init(n);
        for (int[] pair : connections) {
            union(pair[0], pair[1]);
        }
        Set<Integer> set = new HashSet<>();
        for (int i : father) {
            set.add(find(i));
        }
        return set.size() - 1;
    }
    private void init(int n) {
        father = new int[n];
        for (int i = 0; i < n; ++i) {
            father[i] = i;
        }
    }
    private void union(int x, int y) {
        father[x] = find(x);
        father[y] = find(y);
        father[father[x]] = father[y];
    }
    private int find(int x) {
        return father[x] == x ? x : find(father[x]);
    }
}