Algorithms（四）Employee Importance

题目：

Employee Importance

You are given a data structure of employee information, which includes the employee'sunique 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 isnot 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, []]], 1Output: 11Explanation: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.

代码：

/*// Employee infoclass Employee {public:    // It's the unique ID of each node.    // unique id of this employee    int id;    // the importance value of this employee    int importance;    // the id of direct subordinates    vector<int> subordinates;};*/class Solution {public:    int getImportance(vector<Employee*> employees, int id) {        unordered_map<int, Employee*> map;        for(const auto element : employees){            map[element->id] = element;        }        return help(map, id);    }        int help(unordered_map<int, Employee*>& map, const int id){        auto sum = map[id]->importance;        for(const auto element : map[id]->subordinates){            sum += help(map, element);        }        return sum;    }};

算法：

1. 用map容器存储所有employee信息，将id设为Key值，每一个id唯一标识一个employee对象。
2. 对于一个给定的id值，采用深度优先算法，对map[id]的每个子对象递归搜索它们的子对象，直到该子对象为空，返回这条路上的important value的累加值，直到对map[id]的所有子对象都做完上述操作，返回sum最终值。

补充：

jdk 5.0新增了一个遍历叫foreach，冒号前是要遍历的集合，冒号后是实例化一个集合中包含的元素。举例：

ArrayList<类A> list = new ArrayList<类A>();

for(类A : list) {

     操作a;

}