[LeetCode] Employee Importance 员工重要度-白红宇

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, []]], 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.

这道题定义了一种员工类，有id，重要度，和direct report的员工，让我们求某个员工的总重要度。我们要明白的是就算某个员工不直接向你汇报工作，而是向你手下人汇报，这个人的重要度也会算进你的重要度中。这其实就是之前那道的变化形式，我们可以用DFS来做。首先我们想，为了快速的通过id来定位到员工类，需要建立一个id和员工类的映射，然后我们根据给定的员工id来算其重要度。计算方法当然是其本身的重要度加上其所有手下人的重要度，对于手下人，还要累加其手下人的重要度，为了不重复计算某个员工的重要度，我们建立一个集合，将遍历过的员工id放到集合中，这样一旦我们遍历到集合中有的员工，直接返回0即可；否则就将该员工id加入集合中，然后建立一个结果res变量，加上当前员工的重要度，然后遍历其所有手下，对其每个手下人调用递归函数加到res上，最后返回res即可，参见代码如下：

解法一：

public:    int getImportance(vector
     
       employees, int id) {        unordered_set
      
        s;        unordered_map
       
         m;        for (auto e : employees) m[e->id] = e;        return helper(id, m, s);    }    int helper(int id, unordered_map
        
         & m, unordered_set
         
          & s) {        if (s.count(id)) return 0;        s.insert(id);        int res = m[id]->importance;        for (int num : m[id]->subordinates) {            res += helper(num, m, s);        }        return res;    }};

我们也可以用BFS来做，使用一个queue来辅助运算，开始将给定员工id放入，然后当queue不为空进行循环，每次取出队首员工，如果已经访问过了，直接跳过，否则加入集合中，然后累加上当前员工的复杂度到结果res，然后将其所有手下人加入队列等待遍历，参见代码如下：

解法二：

public:    int getImportance(vector
     
       employees, int id) {        int res = 0;        queue
      
        q{
    {id}};        unordered_set
       
         s;        unordered_map
        
          m;        for (auto e : employees) m[e->id] = e;        while (!q.empty()) {            auto t = q.front(); q.pop();            if (s.count(t)) continue;            s.insert(t);            res += m[t]->importance;            for (int num : m[t]->subordinates) {                q.push(num);            }        }        return res;    }};

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