[Lintcode] 629. Minimum Spanning Tree

给出一些Connections，即Connections类，找到一些能够将所有城市都连接起来并且花费最小的边。

如果说可以将所有城市都连接起来，则返回这个连接方法；不然的话返回一个空列表。

例

给出 connections = ["Acity","Bcity",1]，["Acity","Ccity",2]，["Bcity","Ccity",3]，

返回 ["Acity","Bcity",1]，["Acity","Ccity",2]。

class Connection {
	public String city1, city2;
	public int cost;
	public Connection(String city1, String city2, int cost) {
		this.city1 = city1;
		this.city2 = city2;
		this.cost = cost;
	}
}

public class MinimumSpanningTree {
    /**
     * @param connections given a list of connections include two cities and cost
     * @return a list of connections from results
     */
	public List lowestCost(List connections) {
        Collections.sort(connections, comp);
        Map hash = new HashMap();
        int n = 0;
        for (Connection connection : connections) {
            if (!hash.containsKey(connection.city1)) {
                hash.put(connection.city1, ++n);
            }
            if (!hash.containsKey(connection.city2)) {
                hash.put(connection.city2, ++n);
            }
        }

        int[] father = new int[n + 1];

        List results = new ArrayList();
        for (Connection connection : connections) {
            int num1 = hash.get(connection.city1);
            int num2 = hash.get(connection.city2);

            int root1 = find(num1, father);
            int root2 = find(num2, father);
            if (root1 != root2) {
                father[root1] = root2;
                results.add(connection);
            }
        }

        if (results.size() != n - 1) {
            return new ArrayList();
        }
        return results;
    }

    static Comparator comp = new Comparator() {
        public int compare(Connection a, Connection b) {
            if (a.cost != b.cost) {
                return a.cost - b.cost;
            }
            if (a.city1.equals(b.city1)) {
                return a.city2.compareTo(b.city2);
            }
            return a.city1.compareTo(b.city1);
        }
    };

    public int find(int num, int[] father) {
        if (father[num] == 0) {
            return num;
        }

        return father[num] = find(father[num], father);
    }
}