LeetCode [690] Employee Importance

 690. Employee Importance

Easy

You are given a data structure of employee information, which includes the employee's unique id, their importance value and their 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 their 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.

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/* // Definition for Employee. class Employee { public int id; public int importance; public List<Integer> subordinates; }; */ class Solution { Map<Integer, Employee> map; public int getImportance(List<Employee> employees, int id) { map = new HashMap<>(); for(Employee e : employees){ map.put(e.id, e); } return helper(id); } int helper(int id){ int ret = map.get(id).importance; for(int s : map.get(id).subordinates){ ret += helper(s); } return ret; } }

YMSF

写完code之后问了时间空间复杂度，follow-up问如果访问多次应该怎样改code，以及如果不是tree就是普通的graph题应该怎么解

YMSF

但是一开始会问什么样的input data是valid的，讨论了半天，后来说必须是tree structure，然后问充要条件。我一直没想通为什么非得是tree（虽然lc上那题确实有这个条件，但感觉只是为了简单），我觉得只要是无圈图就行。。。然后follow up是如果要频繁调用的话，如何修改提高效率，就是加个记忆化，空间换时间，他说可以，不过改完就没时间了（45分钟，我也不知道为啥不能延长点），看地里说还有follow up是问如果是一般的图怎么做，我一脸问号，不知道有什么区别，按我理解只要有向无圈图总是有唯一子代树的，当时肥肠紧张以至于没敢和他argue，后来肥肠后悔，主要那老哥特别高冷，问什么就答一两个词那种，搞得人心态很差