/*
// Definition for Employee.
class Employee {
public:
int id;
int importance;
vector<int> subordinates;
};
*/
class Solution {
public:
int getImportance(vector<Employee*> employees, int id) {
}
};
690. 员工的重要性
给定一个保存员工信息的数据结构，它包含了员工 唯一的 id ，重要度 和 直系下属的 id 。
比如，员工 1 是员工 2 的领导，员工 2 是员工 3 的领导。他们相应的重要度为 15 , 10 , 5 。那么员工 1 的数据结构是 [1, 15, [2]] ，员工 2的 数据结构是 [2, 10, [3]] ，员工 3 的数据结构是 [3, 5, []] 。注意虽然员工 3 也是员工 1 的一个下属，但是由于 并不是直系 下属，因此没有体现在员工 1 的数据结构中。
现在输入一个公司的所有员工信息，以及单个员工 id ，返回这个员工和他所有下属的重要度之和。
示例：
输入：[[1, 5, [2, 3]], [2, 3, []], [3, 3, []]], 1 输出：11 解释： 员工 1 自身的重要度是 5 ，他有两个直系下属 2 和 3 ，而且 2 和 3 的重要度均为 3 。因此员工 1 的总重要度是 5 + 3 + 3 = 11 。
提示：
相似题目
原站题解
java 解法, 执行用时: 5 ms, 内存消耗: 45.1 MB, 提交时间: 2024-08-26 09:40:22
/*
// Definition for Employee.
class Employee {
public int id;
public int importance;
public List<Integer> subordinates;
};
*/
class Solution {
public int getImportance(List<Employee> employees, int id) {
Map<Integer, Employee> employeeMap = new HashMap<>(employees.size());
for (Employee e : employees) {
employeeMap.put(e.id, e);
}
return dfs(employeeMap, id);
}
private int dfs(Map<Integer, Employee> employeeMap, int id) {
Employee e = employeeMap.get(id);
int res = e.importance;
for (int subId : e.subordinates) {
res += dfs(employeeMap, subId);
}
return res;
}
}
javascript 解法, 执行用时: 60 ms, 内存消耗: 53.2 MB, 提交时间: 2024-08-26 09:40:03
/**
* Definition for Employee.
* function Employee(id, importance, subordinates) {
* this.id = id;
* this.importance = importance;
* this.subordinates = subordinates;
* }
*/
/**
* @param {Employee[]} employees
* @param {number} id
* @return {number}
*/
var GetImportance = function(employees, id) {
const employeeMap = new Map();
for (const e of employees) {
employeeMap.set(e.id, e)
}
function dfs(id) {
const e = employeeMap.get(id);
let res = e.importance;
for (const subId of e.subordinates) {
res += dfs(subId);
}
return res;
}
return dfs(id);
};
cpp 解法, 执行用时: 28 ms, 内存消耗: 19.8 MB, 提交时间: 2024-08-26 09:39:20
/*
// Definition for Employee.
class Employee {
public:
int id;
int importance;
vector<int> subordinates;
};
*/
class Solution {
public:
int getImportance(vector<Employee*>& employees, int id) {
unordered_map<int, Employee*> employee_map;
for (auto e : employees) {
employee_map[e->id] = e;
}
auto dfs = [&](auto&& dfs, int id) -> int {
auto e = employee_map[id];
int res = e->importance;
for (int sub_id : e->subordinates) {
res += dfs(dfs, sub_id);
}
return res;
};
return dfs(dfs, id);
}
};
python3 解法, 执行用时: 126 ms, 内存消耗: 17.7 MB, 提交时间: 2024-08-26 09:38:44
"""
# Definition for Employee.
class Employee:
def __init__(self, id: int, importance: int, subordinates: List[int]):
self.id = id
self.importance = importance
self.subordinates = subordinates
"""
class Solution:
def getImportance(self, employees: List['Employee'], id: int) -> int:
employees = {e.id: e for e in employees}
def dfs(id: int) -> int:
e = employees[id]
return e.importance + sum(dfs(sub) for sub in e.subordinates)
return dfs(id)
python3 解法, 执行用时: 132 ms, 内存消耗: 17.7 MB, 提交时间: 2024-08-26 09:35:04
"""
# Definition for Employee.
class Employee:
def __init__(self, id: int, importance: int, subordinates: List[int]):
self.id = id
self.importance = importance
self.subordinates = subordinates
"""
class Solution:
def getImportance(self, employees: List['Employee'], id: int) -> int:
mp = dict()
total = 0
for e in employees:
mp[e.id] = e
queue = [id]
while len(queue) > 0:
e = mp[queue[0]]
total += e.importance
queue = queue[1:] + e.subordinates
return total
golang 解法, 执行用时: 12 ms, 内存消耗: 6.7 MB, 提交时间: 2021-06-02 10:23:52
/**
* Definition for Employee.
* type Employee struct {
* Id int
* Importance int
* Subordinates []int
* }
*/
func getImportance(employees []*Employee, id int) (total int) {
mp := map[int]*Employee{}
for _, employee := range employees {
mp[employee.Id] = employee
}
// dfs遍历
var dfs func(id int)
dfs = func(id int) {
employee := mp[id]
total += employee.Importance
for _, subId := range employee.Subordinates {
dfs(subId)
}
}
dfs(id)
return total
}
golang 解法, 执行用时: 12 ms, 内存消耗: 6.7 MB, 提交时间: 2021-06-02 10:23:20
/**
* Definition for Employee.
* type Employee struct {
* Id int
* Importance int
* Subordinates []int
* }
*/
func getImportance(employees []*Employee, id int) (total int) {
mp := map[int]*Employee{}
for _, employee := range employees {
mp[employee.Id] = employee
}
// bfs遍历
queue := []int{id}
for len(queue) > 0 {
employee := mp[queue[0]]
total += employee.Importance
queue = append(queue[1:], employee.Subordinates...)
}
return total
}