There are n cities labeled from 1 to n. You are given the integer n and an array connections where connections[i] = [xi, yi, costi] indicates that the cost of connecting city xi and city yi (bidirectional connection) is costi.

Return the minimum cost to connect all the n cities such that there is at least one path between each pair of cities. If it is impossible to connect all the n cities, return -1,

The cost is the sum of the connections' costs used.

 

Example 1:

Input: n = 3, connections = [[1,2,5],[1,3,6],[2,3,1]]
Output: 6
Explanation: Choosing any 2 edges will connect all cities so we choose the minimum 2.

Example 2:

Input: n = 4, connections = [[1,2,3],[3,4,4]]
Output: -1
Explanation: There is no way to connect all cities even if all edges are used.

其实union-find也没有那么难。一个union函数，一个find函数，哦了。
https://leetcode.com/problems/connecting-cities-with-minimum-cost/discuss/344867/Java-Kruskal's-Minimum-Spanning-Tree-Algorithm-with-Union-Find

class Solution {
    
    int[] parent;
    int n;
    
    private void union(int x, int y) {
        int px = find(x);
        int py = find(y);
        
        if (px != py) {
            parent[px] = py;
            n--;
        }
    }
    
    private int find(int x) {
        if (parent[x] == x) {
            return parent[x];
        }
        parent[x] = find(parent[x]); // path compression
        return parent[x];
    }
    
    public int minimumCost(int N, int[][] connections) {
        parent = new int[N + 1];
        n = N;
        for (int i = 0; i <= N; i++) {
            parent[i] = i;
        }
        
        Arrays.sort(connections, (a, b) -> (a[2] - b[2]));
        
        int res = 0;
        
        for (int[] c : connections) {
            int x = c[0], y = c[1];
            if (find(x) != find(y)) {
                res += c[2];
                union(x, y);
            }
        }
        
        return n == 1 ? res : -1;
    }
}