A company has n employees with a unique ID for each employee from 0 to n - 1. The head of the company has is the one with headID.
Each employee has one direct manager given in the manager array where manager[i] is the direct manager of the i-th employee, manager[headID] = -1. Also it's guaranteed that the subordination relationships have a tree structure.
The head of the company wants to inform all the employees of the company of an urgent piece of news. He will inform his direct subordinates and they will inform their subordinates and so on until all employees know about the urgent news.
The i-th employee needs informTime[i] minutes to inform all of his direct subordinates (i.e After informTime[i] minutes, all his direct subordinates can start spreading the news).
Return the number of minutes needed to inform all the employees about the urgent news.
Example 1:
Input: n = 1, headID = 0, manager = [-1], informTime = [0] Output: 0 Explanation: The head of the company is the only employee in the company.
Example 2:
Input: n = 6, headID = 2, manager = [2,2,-1,2,2,2], informTime = [0,0,1,0,0,0] Output: 1 Explanation: The head of the company with id = 2 is the direct manager of all the employees in the company and needs 1 minute to inform them all. The tree structure of the employees in the company is shown.
思路：如果是个tree，那么就是向上返回时间最大值，node加上自己的time加上小弟中的最大值，往上返回；
问题在于怎么建立graph，这题巧妙的是manager是key，value是手下员工list
T: O(N), S: O(N)
class Solution {
public int numOfMinutes(int n, int headID, int[] manager, int[] informTime) {
HashMap<Integer, List<Integer>> graph = new HashMap<>();
int ceo = 0;
for(int i = 0; i < manager.length; i++) {
if(manager[i] == -1) {
ceo = i;
}
graph.putIfAbsent(manager[i], new ArrayList<Integer>());
graph.get(manager[i]).add(i);
}
return dfs(ceo, graph, informTime);
}
private int dfs(int start, HashMap<Integer, List<Integer>> graph, int[] informTime) {
if(!graph.containsKey(start)) {
return 0;
}
int sum = informTime[start];
int levelmax = 0;
for(Integer neighbor: graph.get(start)) {
levelmax = Math.max(levelmax, dfs(neighbor, graph, informTime));
}
sum += levelmax;
return sum;
}
}