标签：ati exit res ems poll nat return seve includes
原题链接在这里：https://leetcode.com/problems/employee-importance/description/
题目：
You are given a data structure of employee information, which includes the employee‘s unique id, his importance value and his directsubordinates‘ id.
For example, employee 1 is the leader of employee 2, and employee 2 is the leader of employee 3. They have importance value 15, 10 and 5, respectively. Then employee 1 has a data structure like [1, 15, [2]], and employee 2 has [2, 10, [3]], and employee 3 has [3, 5, []]. Note that although employee 3 is also a subordinate of employee 1, the relationship is not direct.
Now given the employee information of a company, and an employee id, you need to return the total importance value of this employee and all his subordinates.
Example 1:
Input: [[1, 5, [2, 3]], [2, 3, []], [3, 3, []]], 1 Output: 11 Explanation: Employee 1 has importance value 5, and he has two direct subordinates: employee 2 and employee 3. They both have importance value 3. So the total importance value of employee 1 is 5 + 3 + 3 = 11.
Note:
题解：
类似Clone Graph, Nested List Weight Sum.
可以采用BFS. 每层的Employee挨个加进去.
Time Complexity: O(n). n是root id的所有下属个数, 包括直系下属和非直系下属.
Space: O(n). 有可能都是直系下属.
AC Java:
1 /* 2 // Employee info 3 class Employee { 4 // It‘s the unique id of each node; 5 // unique id of this employee 6 public int id; 7 // the importance value of this employee 8 public int importance; 9 // the id of direct subordinates 10 public List<Integer> subordinates; 11 }; 12 */ 13 class Solution { 14 public int getImportance(List<Employee> employees, int id) { 15 int res = 0; 16 17 HashMap<Integer, Employee> hm = new HashMap<Integer, Employee>(); 18 for(Employee employee : employees){ 19 hm.put(employee.id, employee); 20 } 21 22 LinkedList<Employee> que = new LinkedList<Employee>(); 23 que.add(hm.get(id)); 24 while(!que.isEmpty()){ 25 Employee cur = que.poll(); 26 res += cur.importance; 27 for(int subId : cur.subordinates){ 28 que.add(hm.get(subId)); 29 } 30 } 31 32 return res; 33 } 34 }
DFS.逐层往深dfs.
Time Complexity: O(n). n是root id的所有下属个数, 包括直系下属和非直系下属.
Space: O(n). stack space, 最深的树.
AC Java:
1 /* 2 // Employee info 3 class Employee { 4 // It‘s the unique id of each node; 5 // unique id of this employee 6 public int id; 7 // the importance value of this employee 8 public int importance; 9 // the id of direct subordinates 10 public List<Integer> subordinates; 11 }; 12 */ 13 class Solution { 14 public int getImportance(List<Employee> employees, int id) { 15 HashMap<Integer, Employee> hm = new HashMap<Integer, Employee>(); 16 for(Employee employee : employees){ 17 hm.put(employee.id, employee); 18 } 19 20 return dfs(hm, id); 21 } 22 23 private int dfs(HashMap<Integer, Employee> hm, int id){ 24 int res = 0; 25 Employee cur = hm.get(id); 26 27 res += cur.importance; 28 for(int subId : cur.subordinates){ 29 res += dfs(hm, subId); 30 } 31 return res; 32 } 33 }
标签：ati exit res ems poll nat return seve includes
原文地址：http://www.cnblogs.com/Dylan-Java-NYC/p/7624915.html
