210. Course Schedule II

Medium (Trung bình) C++ Python 🔗 Xem trên LeetCode

📋 Đề Bài

There are a total of numCourses courses you have to take, labeled from 0 to numCourses - 1. You are given an array prerequisites where prerequisites[i] = [a_i, b_i] indicates that you must take course b_i first if you want to take course a_i.

For example, the pair [0, 1], indicates that to take course 0 you have to first take course 1.

Return the ordering of courses you should take to finish all courses. If there are many valid answers, return any of them. If it is impossible to finish all courses, return an empty array.

Example 1:

Input: numCourses = 2, prerequisites = [[1,0]]
Output: [0,1]
Explanation: There are a total of 2 courses to take. To take course 1 you should have finished course 0. So the correct course order is [0,1].

Example 2:

Input: numCourses = 4, prerequisites = [[1,0],[2,0],[3,1],[3,2]]
Output: [0,2,1,3]
Explanation: There are a total of 4 courses to take. To take course 3 you should have finished both courses 1 and 2. Both courses 1 and 2 should be taken after you finished course 0.
So one correct course order is [0,1,2,3]. Another correct ordering is [0,2,1,3].

Example 3:

Input: numCourses = 1, prerequisites = []
Output: [0]

Constraints:

1 <= numCourses <= 2000
0 <= prerequisites.length <= numCourses * (numCourses - 1)
prerequisites[i].length == 2
0 <= a_i, b_i < numCourses
a_i != b_i
All the pairs [a_i, b_i] are distinct.

🧠 Thuật Toán & Kỹ Thuật

DFS (Tìm kiếm theo chiều sâu)Hash Table (Bảng băm)Graph (Đồ thị)Bit Manipulation (Thao tác bit)

⏱️ Thời gian O(V+E)

💾 Không gian O(V)

💻 Lời Giải

C++ 0210-course-schedule-ii.cpp

class Solution {
    map<int, vector<int>> adj;
    vector<int> color;
    vector<int> ans;

public:
    bool dfs(int u) {
        color[u] = 1;

        for (int v : adj[u]) {
            if (color[v] == 0 && !dfs(v)) {
                return false;
            }
            else if (color[v] == 1) {
                return false;
            }
        }

        ans.push_back(u);
        color[u] = 2;
        return true;
    }

    vector<int> findOrder(int numCourses, vector<vector<int>>& prerequisites) {
        color.resize(numCourses);
        const int n = numCourses;

        for (vector<int> p : prerequisites) {
            int u = p[0], v = p[1];
            adj[u].push_back(v);
        }

        for (int i = 0; i < numCourses; ++i) {
            if (color[i] == 0 && !dfs(i)) {
                return {};
            }
        }

        return ans;
    }
};

Python 0210-course-schedule-ii.py

class Solution:
    def findOrder(self, numCourses: int, prerequisites: List[List[int]]) -> List[int]:
        adj = defaultdict(list)
        
        for u, v in prerequisites:
            adj[u].append(v)
        
        color = [0] * numCourses
        topo = []
        
        def dfs(u: int) -> bool:
            if color[u] == 1:
                return False
            color[u] = 1
            for v in adj[u]:
                if color[v] == 0:
                    if not dfs(v):
                        return False
                elif color[v] == 1:
                    return False
            color[u] = 2
            topo.append(u)
            return True
        
        for u in range(0, numCourses):
            if color[u] == 0 and not dfs(u):
                return []
        
        return topo