Breadth-first Search -- Leetcode problem690. Employee Importance

描述：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.

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：这道题看起来题目描述十分复杂，其实并不困难，关键在于找到一个合适的容器将题目中所给的数据装入容器之中，然后的工作就是进行BFS或DFS查找。

思路一：使用unordered_map作为容器，用BFS算法进行搜索。

/*
// Employee info
class 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*> my_map;
for (auto e : employees)
my_map[e -> id] = e;
int sum = 0;
queue<int> my_queue;
my_queue.push(id);
while (!my_queue.empty()) {
sum += my_map[my_queue.front()] -> importance;
vector<int> temp = my_map[my_queue.front()] -> subordinates;
my_queue.pop();
for (auto i : temp)
my_queue.push(i);
}
return sum;
}
};

思路二：用unordered-map做容器，用递归求解

class Solution {
public:
int getImportance(vector<Employee*> employees, int id) {
unordered_map<int, Employee*> my_map;
for (Employee* e: employees) {
my_map[e -> id] = e;
}
return findsum(my_map, id);
}

int findsum(unordered_map<int, Employee*> &my_map, int id) {
int sum = my_map[id] -> importance;
for (int i:my_map[id] -> subordinates) {
sum += findsum(my_map, i);
}
return sum;
}
};