LeetCode.690 员工的重要性

最新推荐文章于 2025-05-27 01:15:00 发布

原创最新推荐文章于 2025-05-27 01:15:00 发布 · 97 阅读

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#leetcode #stack #递归算法

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本文介绍了一种通过递归或栈实现的方法来计算组织结构中员工的重要性总和。该方法首先将所有员工信息存储到哈希表中，接着利用栈进行深度优先搜索遍历员工及其下属，累计计算他们的总重要性。

原题

https://leetcode-cn.com/problems/employee-importance/

在这里插入图片描述

思路

使用map转存数据
递归/栈

题解

/*
// Definition for Employee.
class Employee {
    public int id;
    public int importance;
    public List<Integer> subordinates;
};
*/

class Solution {
    public int getImportance(List<Employee> employees, int id) {
        int res = 0;

        Map<Integer, Employee> map = new HashMap<Integer, Employee>();
        for (Employee employee : employees) {
            map.put(employee.id, employee);
        }

        Stack<Integer> stack = new Stack<>();
        Employee employee = map.get(id);
        List<Integer> children = employee.subordinates;
        res += employee.importance;
        for (Integer child : children) {
            stack.push(child);
        }

        while(!stack.isEmpty()){
            Employee emp = map.get(stack.pop());
            res += emp.importance;
            List<Integer> subordinates = emp.subordinates;
            for (Integer child : subordinates) {
                stack.push(child);
            }
        }

        return res;
    }
}

/*
// Definition for Employee.
class Employee {
    public int id;
    public int importance;
    public List<Integer> subordinates;
};
*/

class Solution {
    public int getImportance(List<Employee> employees, int id) {
        Map<Integer, Employee> map = new HashMap<Integer, Employee>();
        for (Employee employee : employees) {
            map.put(employee.id, employee);
        }

        return DFS(map, id);
    }

    private int DFS(Map<Integer, Employee> map, Integer id){
        Employee employee = map.get(id);
        List<Integer> subordinates = employee.subordinates;
        for (Integer child : subordinates) {
            employee.importance += DFS(map, child);
        }
        return employee.importance;
    }

}