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  • leetcode 690. Employee Importance——本质上就是tree的DFS和BFS

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

    经典BFS:tree的层序遍历思想

    """
    # 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
            """
            """
            Input: [[1, 5, [2, 3]], [2, 3, []], [3, 3, []]], 1
            Output: 11
            use dict to find its sub employee
            """
            emp_dict = {e.id:e for i,e in enumerate(employees)}
            root = emp_dict[id]
            # tree BFS
            q = [root]
            ans = 0
            while q:
                q2 = []
                for node in q:
                    ans += node.importance
                    for i in node.subordinates:
                        q2.append(emp_dict[i])
                q = q2
            return ans        

    DFS:

    """
    # 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
            """
            """
            Input: [[1, 5, [2, 3]], [2, 3, []], [3, 3, []]], 1
            Output: 11
            use dict to find its sub employee
            """
            emp_dict = {e.id:e for i,e in enumerate(employees)}
            root = emp_dict[id]
            
            # tree DFS 
            def dfs(root, emp_dict):            
                score = root.importance
                for i in root.subordinates:
                    score += dfs(emp_dict[i], emp_dict)
                return score
                
            return dfs(root, emp_dict)    
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  • 原文地址:https://www.cnblogs.com/bonelee/p/8570877.html
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