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690_Employee_Importance.java
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690_Employee_Importance.java
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// 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:
//
// One employee has at most one direct leader and may have several subordinates.
// The maximum number of employees won't exceed 2000.
//
//
/*
// 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 {
Map<Integer, Employee> emap;
public int getImportance(List<Employee> employees, int queryid) {
emap = new HashMap();
for (Employee e: employees) emap.put(e.id, e);
Stack <Employee> stack=new Stack<>();
int sum=0;
stack.push(emap.get(queryid));
while(!stack.empty()){
Employee tmp=stack.peek();
sum+=tmp.importance;
stack.pop();
for(int i=tmp.subordinates.size()-1;i>=0;i--){
stack.push(emap.get(tmp.subordinates.get(i)));
}
}
return sum;
}
/*
public int dfs(int eid) {
Employee employee = emap.get(eid);
int ans = employee.importance;
for (Integer subid: employee.subordinates)
ans += dfs(subid);
return ans;
}
*/
}