[LeetCode] 690. Employee Importance_Easy tag: BFS

标签：oop   input   queue   seve   includes   most   unique   结束   import   

You are given a data structure of employee information, which includes the employee‘s unique id, his importance value and his directsubordinates‘ 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:

1. One employee has at most one direct leader and may have several subordinates.
2. The maximum number of employees won‘t exceed 2000.

这个题目很典型的DFS或者BFS, 这里我用BFS, 只是需要注意的一点是input 是employee 为data structure, 但是只需要用两个dictionary来分别存相对应的数值就好, 换汤不换药. 本质还是BFS.

1. constrains:

    1) input 的max size 为2000, 可以为[]

    2) edge: maybe id not in employees. 此时, 直接return ans(init:0)

2. ideas:

     BFS:      T: O(n)  # n 为number of employees

                   S: O(n) # 2n dictionary, n queue, so total is O(n)

    1) d1, d2, ans 分别存importance, subordinates, 0

    2) if id not in d1, return ans   # edge

    3) queue (init: [id]), visited(init: set())

   4) while queue: new_id = queue.popleft(), ans += new_id.importance, visited.add(new_id)

   5) for each in new_id.subordinates: 判断是否visited过, 如果没有, queue.append(each)

   6) 结束while loop, return ans

3. code

 1 class Solution:
 2     def getImportance(self, employees, id):
 3         d1, d2, ans = {}, {}, 0
 4         for e in employees:
 5             d1[e.id] = e.importance
 6             d2[e.id] = e.subordinates
 7         if id not in d1: return ans
 8         queue, visited = collections.deque([id]), set()
 9         while queue:
10             new_id = queue.popleft()
11             ans += d1[new_id]
12             visited.add(new_id)
13             for each in d2[new_id]:
14                 if each not in visited and each in d1: # add each in d1 incase input [[1,5, [2]]] like this, but test cases dont include, 多思考点不是坏处
15                     queue.append(each)
16         return ans

4. test cases

  1) [], 1   

  2) [[1, 4, []]], 2

  3) [[1, 5, [2, 3]], [2, 3, []], [3, 3, []]], 1

[LeetCode] 690. Employee Importance_Easy tag: BFS

标签：oop   input   queue   seve   includes   most   unique   结束   import   

原文地址：https://www.cnblogs.com/Johnsonxiong/p/9256344.html