Leetcode刷题 690 Employee Importance

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.

个人答案：

# Employee info
class Employee(object):
    def __init__(self, id, importance, subordinates):
        # It's the unique id of each node.
        # unique id of this employee
        self.id = id
        # the importance value of this employee
        self.importance = importance
        # the id of direct subordinates
        self.subordinates = subordinates

class Solution(object):

    def getImportance(self, employees, id):
        """
        :type employees: Employee
        :type id: int
        :rtype: int
        """

        id_list = []
        importance_list = []
        sub_list = []
        for i in range(0, len(employees)):
            id_list.append(employees[i].id)
            importance_list.append(employees[i].importance)
            sub_list.append(employees[i].subordinates)

        id_importance_zipped = zip(id_list, importance_list)
        id_sub_zipped = zip(id_list, sub_list)
        id_importance_dict = dict(id_importance_zipped)  # 全局字典，就是hashtable
        id_sub_dict = dict(id_sub_zipped)  # 全局字典


        def dfs(id):
            total_importance = id_importance_dict[id]
            if len(id_sub_dict[id]) == 0:  # 若是没下属
                return total_importance
            else:#若是有下属
                for sub_id in id_sub_dict[id]:
                    total_importance = total_importance + dfs(sub_id)
                return total_importance
        return dfs(id)



sol=Solution()
emp1=Employee(1,5,[2,3])
emp2=Employee(2,3,[4])
emp3=Employee(3,4,[])
emp4=Employee(4,1,[])
employees=[emp1,emp2,emp3,emp4]
id=1
total=sol.getImportance(employees,id)

total
Out[1]: 13

说实话我并非一遍写对的，在逐步探索的过程当中我意识到：

1.我选了一道tag为HashTable的这道题，从网上查了一下python经过字典实现hashtable，茅塞顿开。

    想分别把id-imporatnce这个键值对

                和id-subordinates这个键值对 

   存成两个字典 

2. 我意识到上述两个字典必须是全局变量，不然的话，好比本题中我举的例子：

id=1的员工有子员工id=2和id=3 ，而id=2的员工有子员工id=4, 若是上述两个字典不是全局变量，则id=4没法查找到键为4的键值对，会报Keyerror的错误

3.题意确实是深度优先搜索

综上所述写出了正确答案，可是超级慢399ms只战胜了6.98%的用户。

因而看了别人更快的代码（172ms,最快）：

class Solution(object):
    def getImportance(self, employees, id):
        """
        :type employees: Employee
        :type id: int
        :rtype: int
        """
        return employees[id - 1].importance + sum(
            [self.getImportance(employees, subordinate_id) for subordinate_id in employees[id - 1].subordinates])

比这个次之的是discussion里一个与我写法更接近，更易于理解的方法：

class Employee(object):
    def __init__(self, id, importance, subordinates):
        # It's the unique id of each node.
        # unique id of this employee
        self.id = id
        # the importance value of this employee
        self.importance = importance
        # the id of direct subordinates
        self.subordinates = subordinates


class Solution(object):

    def getImportance(self, employees, id):
        """
        :type employees: Employee
        :type id: int
        :rtype: int
        """
        # Time: O(n)
        # Space: O(n)
        emps = {employee.id: employee for employee in employees}
        def dfs(id):
            subordinates_importance = sum([dfs(sub_id) for sub_id in emps[id].subordinates])
            return subordinates_importance + emps[id].importance
        return dfs(id)

sol=Solution()
#employees=[[1, 5, [2, 3]], [2, 3, []], [3, 3, []]]
emp1=Employee(1,5,[2,3])
emp2=Employee(2,3,[4])
emp3=Employee(3,4,[])
emp4=Employee(4,1,[])
employees=[emp1,emp2,emp3,emp4]
id=1
total=sol.getImportance(employees,id)

这个方法设置字典的方法很巧妙，把id与employee对象作成键值对，这样就没必要像我同样构造id-importance,id-subordinates两个字典了

经验总结：

1. 我一开始把结果sum=sol.getImportance(employees,id)

这样的话我再调试的时候，写sum([6,7,8])

就会报TypeError: 'int' object is not callable

发现是个人变量名占用了关键字的名字（也不能占用函数的名字）

深入教训，不能再犯

2.

sum([ ])
Out[4]: 0
sum([id for id in []])
Out[5]: 0