690. Employee Importance(HashTable)

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.

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.

首先复习一下HashMap和Queue的常用方法

HashMap
SN  方法及描述
1   void clear(): 移除所有映射关系
2   Object clone(): 返回此HashMap实例的浅表副本：键和值本身不被复制。
3   boolean containsKey(Object key): 判断是否包含键
4   boolean containsValue(Object value) : 判断是否包含值
5   int size(): 返回数量
6   Object get(Object key): 返回指定键的值或者null
7   boolean isEmpty(): 判断是否为空
8   Set keySet(): 返回键的set
9   Object put(Object key, Object value): 关联与指定键的指定的值
10  putAll(Map m): 复制所有指定映射到此映射，这些映射关系将替换所有当前映射中键的映射关系
11  Object remove(Object key): 移除映射
12  Collection values(): 返回值的collection

Queue
队列是一种特殊的线性表，它只允许在表的前端进行删除操作，而在表的后端进行插入操作。
LinkedList类实现了Queue接口，因此我们可以把LinkedList当成Queue来用。
1    offer(): 添加元素
2    poll(): 返回第一个元素，并在队列中删除
3    element(): 返回第一个元素
4    peek(): 返回第一个元素

最简单做法

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

另一种思路，用hashMap将id与其对应的数据结构映射起来，可以通过两种方法去解，分别是BFS和DFS

BFS

/*
// 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 subordinates;
};
*/
class Solution {
    public int getImportance(List employees, int id) {
        int total = 0;
        Map map = new HashMap<>();
        for(Employee employee: employees){
            map.put(employee.id, employee);
        }
        Queue queue = new LinkedList<>();
        queue.offer(map.get(id));
        while(!queue.isEmpty()){
            Employee current = queue.poll();
            total += current.importance;
            for (int subordinate: current.subordinates){
                queue.offer(map.get(subordinate));
            }
        }
        return total;
    }
}

DFS

class Solution {
    public int getImportance(List employees, int id) {
        Map map = new HashMap();
        for(int i=0;i map,int i){
        Employee leader = map.get(i);
        int importance = leader.importance;
        List subordinates = leader.subordinates;
        for(Integer j:subordinates)
            importance = importance + getSubImportance(map,j);
        return importance;  
    }                                                   
}