690. Employee Importance - Easy

最新推荐文章于 2021-01-05 18:20:08 发布

转载最新推荐文章于 2021-01-05 18:20:08 发布 · 65 阅读

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原文链接：http://www.cnblogs.com/fatttcat/p/10142634.html

本文介绍了一种员工信息的数据结构，包括员工的唯一ID、重要性值及其直接下属的ID。通过实例展示了如何计算一名员工及其所有下属的总重要性值，使用哈希映射提高查找效率，实现时间复杂度O(n)和空间复杂度O(n)。

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:

One employee has at most one direct leader and may have several subordinates.
The maximum number of employees won't exceed 2000.

用hashmap存（id, employee），以方便查找

time: O(n), space: O(n)

/*
// 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) {
        HashMap<Integer, Employee> map = new HashMap<>();
        for(Employee e : employees) {
            map.put(e.id, e);
        }
        return dfs(map, id, map.get(id).importance);
    }
    
    private int dfs(HashMap<Integer, Employee> map, int id, int sum) {
        if(map.get(id).subordinates.size() == 0)
            return sum;
        for(int sub : map.get(id).subordinates) {
            sum += dfs(map, sub, map.get(sub).importance);
        }
        return sum;
    }
}

转载于:https://www.cnblogs.com/fatttcat/p/10142634.html