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  • 690. Employee Importance员工权限重要性

    [抄题]:

    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.

     [暴力解法]:

    时间分析:

    空间分析:

     [优化后]:

    时间分析:

    空间分析:

    [奇葩输出条件]:

    [奇葩corner case]:

    [思维问题]:

    [一句话思路]:

    调用Employee root等自定义的新型数据结构,只需要加点调用就行了

    [输入量]:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入):

    [画图]:

    [一刷]:

    1. subordinate就是 int型ID,直接用就行

    [二刷]:

    [三刷]:

    [四刷]:

    [五刷]:

      [五分钟肉眼debug的结果]:

    [总结]:

    [复杂度]:Time complexity: O(n) Space complexity: O(n)

    [英文数据结构或算法,为什么不用别的数据结构或算法]:

    求出“所有”:DFS嵌套

    [关键模板化代码]:

    d f s 一直相加:

    public int getImportanceHelper(Map<Integer, Employee> map, int rootId) {
            //ini : res
            Employee root = map.get(rootId);
            int res = root.importance;
            
            //add all subordinates
            for (int subordinate : root.subordinates) {
                res += getImportanceHelper(map, subordinate);
            }
            
            //return res
            return res;
        }

    [其他解法]:

    [Follow Up]:

    [LC给出的题目变变变]:

     [代码风格] :

    /*
    // Employee info
    class Employee {
        // It's the unique id of each node;
        // unique id of this employee
        public int id;
        // the importance value of this employee
        public int importance;
        // the id of direct subordinates
        public List<Integer> subordinates;
    };
    */
    class Solution {
        public int getImportance(List<Employee> employees, int id) {
            //ini: map
            HashMap<Integer, Employee> map = new HashMap<Integer, Employee>();
            
            //put all into map
            for (Employee employee : employees) {
                map.put(employee.id, employee);
            }
            
            //call helper
            return getImportanceHelper(map, id);
        }
        
        public int getImportanceHelper(Map<Integer, Employee> map, int rootId) {
            //ini : res
            Employee root = map.get(rootId);
            int res = root.importance;
            
            //add all subordinates
            for (int subordinate : root.subordinates) {
                res += getImportanceHelper(map, subordinate);
            }
            
            //return res
            return res;
        }
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/immiao0319/p/8933994.html
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