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:
- One employee has at most one direct leader and may have several subordinates.
- 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
原文地址：https://www.cnblogs.com/Johnsonxiong/p/9256344.html