leetcode-690-Employee-Importance

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描述

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:

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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:

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

分析

题目大意是给出一些员工信息，包括员工 id，重要度，及其直接下属，要求根据员工 id，求出他以及他的所有下属（包括下属的下属等等）的重要度的和。

可以采用深度优先遍历或广度优先遍历。

解决方案1(Java)

深度优先遍历。

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/*
// 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, List<Integer>> subordinatesMap = new HashMap<>();
        HashMap<Integer, Integer> importanceMap = new HashMap<>();
        for (Employee employee: employees) {
            subordinatesMap.put(employee.id, employee.subordinates);
            importanceMap.put(employee.id, employee.importance);
        }
        return dfs(employees, subordinatesMap, importanceMap, id);
    }
    
    private static int dfs(List<Employee> employees, HashMap<Integer, List<Integer>> subordinatesMap, 
                           HashMap<Integer, Integer> importanceMap, int id) {
        int sum = 0;
        List<Integer> subordinates = subordinatesMap.get(id);
        sum += importanceMap.get(id);
        for (int subordinate: subordinates) {
            sum += dfs(employees, subordinatesMap, importanceMap, subordinate);
        }
        return sum;
    }
}

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题目来源