Leetcode 690. Employee Importance(python+cpp)

Leetcode Employee Importance

• 題目
• 解法1：BFS
• 解法2：DFS
• CPP+DFS版本解法

題目

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.

解法1：BFS

class Solution:
    def getImportance(self, employees: List['Employee'], id: int) -> int:
        score = 0
        emap = {e.id: e for e in employees}
        q = collections.deque()
        q.append(id)
        while q:
            curr = q.popleft()
            score += emap[curr].importance
            subs = emap[curr].subordinates
            if len(subs)>0:
                for sub in subs:
                    q.append(sub)
        return score
           

時間複雜度：O(N)

空間複雜度：O(N)

解法2：DFS

兩種DFS的方法其實一樣，隻是提供兩種不同表示方法

class Solution:
    def getImportance(self, employees: List['Employee'], id: int) -> int:
        emap = {e.id: e for e in employees}
        def helper(eid):
            curr = emap[eid]
            return (curr.importance + sum(helper(eid) for eid in curr.subordinates))
        return helper(id)
        
        emap = {e.id: e for e in employees}
        def helper(eid):
            nonlocal score
            curr = emap[eid]
            score += curr.importance
            for eid in curr.subordinates:
                helper(eid)
        score = 0
        helper(id)
        return score
           

時間複雜度：O(N)

空間複雜度：O(N)

CPP+DFS版本解法

class Solution {
public:
    int score;
    int getImportance(vector<Employee*> employees, int id) {
        unordered_map<int, Employee*> map;
        for(auto e:employees){
            map[e->id] = e;
        }
        score = 0;
        helper(id,map);
        return score;
    }
    void helper(int eid, unordered_map<int, Employee*>& map){
            // Employee* curr
            // curr = map[eid];
            score = score + map[eid]->importance;
            for(auto eid:map[eid]->subordinates){
                helper(eid,map);
            }
        }
};