LeetCode#690: Employee Importance

Description

You are given a data structure of employee information, which includes the employee’s unique id, his importance value and his direct subordinates’ 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.

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;
};

Example

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.

Solution

这是一道较简单的深度优先搜索（DFS）的题。我们需要根据指定的员工id找到归他管的员工id，再根据归他管的员工id继续递归地向下查找，并将他们的importance值相加起来。

而整道题的核心就在于如何根据id找到一个员工。如果假设员工id在列表中是从1-n顺序存放的，正如上面示例那样，那么可以直接使用深度优先搜索：

class Solution {
    public int getImportance(List<Employee> employees, int id) {
    	Employee employee = employees.get(id-1);
    	int importance = employee.importance;
    	for(Integer subId : employee.subordinates) {
    		importance += getImportance(employees, subId);
    	}
    	return importance;
    }
}

但是此题的员工id在List中是随机存放的，因此我们要先使用一个HashMap保存每个id对应的员工对象，再进行深度优先搜索：

class Solution {
    public int getImportance(List<Employee> employees, int id) {
        Map<Integer, Employee> map = new HashMap<>();
        for(Employee employee : employees)
        	map.put(employee.id, employee);
        return getImportanceCore(map, id);
    }
    
    public int getImportanceCore(Map<Integer, Employee> map, int id) {
    	Employee employee = map.get(id);
    	int importance = employee.importance;
    	for(Integer subId : employee.subordinates) {
    		importance += getImportanceCore(map, subId);
    	}
    	return importance;
    }
}