标签：code tree array sorted rip bsp rom ted sam
Given a list of Connections, which is the Connection class (the city name at both ends of the edge and a cost between them), find edges that can connect all the cities and spend the least amount.
Return the connects if can connect all the cities, otherwise return empty list.
Example 1:
Input:
["Acity","Bcity",1]
["Acity","Ccity",2]
["Bcity","Ccity",3]
Output:
["Acity","Bcity",1]
["Acity","Ccity",2]
Example 2:
Input:
["Acity","Bcity",2]
["Bcity","Dcity",5]
["Acity","Dcity",4]
["Ccity","Ecity",1]
Output:
[]
Explanation:
No way
Return the connections sorted by the cost, or sorted city1 name if their cost is same, or sorted city2 if their city1 name is also same.
首先学习一下并查集的一些概念：
Disjoint Set: 对于Set里面的Node，只能属于一个Set。
支持两种操作：find（找到这个Node属于哪个Set）和 union（合并两个Set）
树是一种特例：
- Set指的是Root of the tree。这也很好理解，如果Node在一棵树上，那么他们share同一个树的Root。
- 复杂度：find(o(h)) - h means the height of the tree，union(o(1))
Reference：https://www.youtube.com/watch?v=UBY4sF86KEY
对于本道题（谁背谁会题），解法是先按照Cost Sort Connections。然后遍历Connections，查和并。如果两个node不属于一个Set，push到result list里。返回的时候检查一下result和点数目的关系（如果全部联通的话，Edge数应为Node数减1）。
代码如下。可以进一步优化：由于find的复杂度为树的高度，那么并的时候以更高的树的Root为Parent则可以减少find的复杂度。
1 /** 2 * Definition for a Connection. 3 * public class Connection { 4 * public String city1, city2; 5 * public int cost; 6 * public Connection(String city1, String city2, int cost) { 7 * this.city1 = city1; 8 * this.city2 = city2; 9 * this.cost = cost; 10 * } 11 * } 12 */ 13 public class Solution { 14 /** 15 * @param connections given a list of connections include two cities and cost 16 * @return a list of connections from results 17 */ 18 Map<String, String> parents = new HashMap<>(); 19 20 public List<Connection> lowestCost(List<Connection> connections) { 21 Collections.sort(connections, new MyComparator()); 22 23 List<Connection> res = new ArrayList<>(); 24 Set<String> nodes = new HashSet<>(); 25 26 for (Connection c : connections) { 27 nodes.add(c.city1); 28 nodes.add(c.city2); 29 30 String s1 = find(c.city1); 31 String s2 = find(c.city2); 32 33 if (!s1.equals(s2)) { 34 union(s1, s2); 35 res.add(c); 36 } 37 } 38 39 if (nodes.size() != (res.size() + 1)) { 40 return new ArrayList<Connection>(); 41 } 42 43 return res; 44 } 45 46 private String find(String x) { 47 String curr = x; 48 49 while (parents.get(curr) != null) { 50 curr = parents.get(curr); 51 } 52 53 return curr; 54 } 55 56 private void union(String s1, String s2) { 57 parents.put(s1, s2); 58 } 59 60 class MyComparator implements Comparator<Connection> { 61 @Override 62 public int compare(Connection a, Connection b) { 63 if (a.cost != b.cost) { 64 return a.cost - b.cost; 65 } 66 67 if (!a.city1.equals(b.city1)) { 68 return a.city1.compareTo(b.city1); 69 } 70 71 return a.city2.compareTo(b.city2); 72 } 73 } 74 }
[LeetCode] Minimum Spanning Tree
标签：code tree array sorted rip bsp rom ted sam
原文地址：https://www.cnblogs.com/yinger33/p/10943505.html
