leetcode 690. Employee importance
leetcode 690. Employee importance
題目:
You are given a data structure of employee information, which includes the employee’s unique id, his importance valueand 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:
- One employee has at most one direct leader and may have several subordinates.
- The maximum number of employees won’t exceed 2000.
解法:
這個題大致意思是,給定一個id我們找出所有他的下屬,並對其所有下屬、下屬的下屬以及他自己的importance求和,最後返回求和結果。
具體做的話,可以看作是對一顆二叉樹求層次遍歷,對每一個結點都進行求和,將子結點的和返回給父節點。
為了降低時間複雜度,我們可以把 list 中的結點存放在 hashmap 中.
程式碼:
/*
// 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<Integer> subordinates;
};
*/
class Solution {
public int getImportance(List<Employee> employees, int id) {
if (employees == null || employees.size() == 0) {
return 0;
}
Map<Integer, Employee> map = new HashMap<>();
for (Employee e : employees) {
map.put(e.id, e);
}
return helper(map, id);
}
private int helper(Map<Integer, Employee> map, int id) {
if (!map.containsKey(id)) {
return 0;
}
int total = map.get(id).importance;
for (int subId : map.get(id).subordinates) {
total += helper(map, subId);
}
return total;
}
};