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;
    }
}
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原文地址:https://www.cnblogs.com/immiao0319/p/8933994.html