[抄题]:
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.
[暴力解法]:
时间分析:
空间分析:
[优化后]:
时间分析:
空间分析:
[奇葩输出条件]:
[奇葩corner case]:
[思维问题]:
[一句话思路]:
调用Employee root等自定义的新型数据结构,只需要加点调用就行了
[输入量]:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入):
[画图]:
[一刷]:
- subordinate就是 int型ID,直接用就行
[二刷]:
[三刷]:
[四刷]:
[五刷]:
[五分钟肉眼debug的结果]:
[总结]:
[复杂度]:Time complexity: O(n) Space complexity: O(n)
[英文数据结构或算法,为什么不用别的数据结构或算法]:
求出“所有”:DFS嵌套
[关键模板化代码]:
d f s 一直相加:
public int getImportanceHelper(Map<Integer, Employee> map, int rootId) { //ini : res Employee root = map.get(rootId); int res = root.importance; //add all subordinates for (int subordinate : root.subordinates) { res += getImportanceHelper(map, subordinate); } //return res return res; }
[其他解法]:
[Follow Up]:
[LC给出的题目变变变]:
[代码风格] :
/* // Employee info class Employee { // It's the unique id of each node; // unique id of this employee public int id; // the importance value of this employee public int importance; // the id of direct subordinates public List<Integer> subordinates; }; */ class Solution { public int getImportance(List<Employee> employees, int id) { //ini: map HashMap<Integer, Employee> map = new HashMap<Integer, Employee>(); //put all into map for (Employee employee : employees) { map.put(employee.id, employee); } //call helper return getImportanceHelper(map, id); } public int getImportanceHelper(Map<Integer, Employee> map, int rootId) { //ini : res Employee root = map.get(rootId); int res = root.importance; //add all subordinates for (int subordinate : root.subordinates) { res += getImportanceHelper(map, subordinate); } //return res return res; } }