题目

题目链接：
https://www.lintcode.com/problem/629/

思路

核心1：对connections进行排序，根据开销升序排序
核心2：并查集，合并集合，记录下合并的边缘
核心3：如果合并完后，集合数是一个，说明是一棵树，返回边缘集合，否则发你空集合

Java答案

/**
 * Definition for a Connection.
 * public 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 Solution {
    /**
     * @param connections given a list of connections include two cities and cost
     * @return a list of connections from results
     */
    public List<Connection> lowestCost(List<Connection> connections) {
        //核心1：排序，根据开销排序
           Collections.sort(connections,(a,b)->{
                if(a.cost == b.cost){
                    if(a.city1.equals(b.city1)){
                        return a.city2.compareTo(b.city2);
                    }else{
                        return a.city1.compareTo(b.city1);
                    }
                }else{
                    return a.cost-b.cost;
                }
            });
            UF uf = new UF();
            List<Connection> ans = new ArrayList<>();
            int mincost=0; //本题中没用，统计最小生成树的最小花开销，比如牛客上相同的题目就是求开销
            for (Connection v : connections) {
                String f = v.city1;
                String t = v.city2;
                
                String f1 = uf.finx(f);
                String f2 = uf.finx(t);
                
                if(!f1.equals(f2)){
                    ans.add(v);
                    mincost+= v.cost;
                    
                    uf.union(f,t);
                }
            }
            
            if(uf.sets==1){ //集合不止一个，说明不能连通
                return ans;
            }else{
                return new ArrayList<>();
            }
        }

        static class UF{
            public Map<String,String> parents = new HashMap<>();
            public Map<String,Integer> size = new HashMap<>();
            public Map<Integer,String> help = new HashMap<>();

            int sets =0;

            public  String finx(String x){
                if(!parents.containsKey(x)){
                    parents.put(x,x);
                    size.put(x,1);
                    sets++;
                }

                int hi=0;
                while (!x.equals(parents.get(x))){
                    help.put(hi++,x);
                    x= parents.get(x);
                }

                for(hi--;hi>=0;hi--){
                    parents.put(help.get(hi),x);
                }

                return x;
            }

            public void union(String a,String b){
                String f1 = finx(a);
                String f2 = finx(b);

                if(!f1.equals(f2)){
                    int s1 = size.get(f1);
                    int s2 = size.get(f2);

                    int sum = s1+s2;
                    if(s1>=s2){
                        size.put(f1,sum);
                        parents.put(f2,f1);
                    }else{
                        size.put(f2,sum);
                        parents.put(f1,f2);
                    }

                    sets--;
                }
            }

        }
}