690. Employee Importance员工权限重要性

［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］：

1. 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;
    }
}