Breadth-first Search -- Leetcode problem690. Employee Importance
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- 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 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*> 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;}};
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