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Medium |
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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] = [ai, bi] indicates that you must take course bi first if you want to take course ai.
- For example, the pair
[0, 1], indicates that to take course0you have to first take course1.
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 <= 20000 <= prerequisites.length <= numCourses * (numCourses - 1)prerequisites[i].length == 20 <= ai, bi < numCoursesai != bi- All the pairs
[ai, bi]are distinct.
class Solution:
def findOrder(self, numCourses: int, prerequisites: List[List[int]]) -> List[int]:
g = defaultdict(list)
indeg = [0] * numCourses
for a, b in prerequisites:
g[b].append(a)
indeg[a] += 1
ans = []
q = deque(i for i, x in enumerate(indeg) if x == 0)
while q:
i = q.popleft()
ans.append(i)
for j in g[i]:
indeg[j] -= 1
if indeg[j] == 0:
q.append(j)
return ans if len(ans) == numCourses else []class Solution {
public int[] findOrder(int numCourses, int[][] prerequisites) {
List<Integer>[] g = new List[numCourses];
Arrays.setAll(g, k -> new ArrayList<>());
int[] indeg = new int[numCourses];
for (var p : prerequisites) {
int a = p[0], b = p[1];
g[b].add(a);
++indeg[a];
}
Deque<Integer> q = new ArrayDeque<>();
for (int i = 0; i < numCourses; ++i) {
if (indeg[i] == 0) {
q.offer(i);
}
}
int[] ans = new int[numCourses];
int cnt = 0;
while (!q.isEmpty()) {
int i = q.poll();
ans[cnt++] = i;
for (int j : g[i]) {
if (--indeg[j] == 0) {
q.offer(j);
}
}
}
return cnt == numCourses ? ans : new int[0];
}
}class Solution {
public:
vector<int> findOrder(int numCourses, vector<vector<int>>& prerequisites) {
vector<vector<int>> g(numCourses);
vector<int> indeg(numCourses);
for (auto& p : prerequisites) {
int a = p[0], b = p[1];
g[b].push_back(a);
++indeg[a];
}
queue<int> q;
for (int i = 0; i < numCourses; ++i) {
if (indeg[i] == 0) {
q.push(i);
}
}
vector<int> ans;
while (!q.empty()) {
int i = q.front();
q.pop();
ans.push_back(i);
for (int j : g[i]) {
if (--indeg[j] == 0) {
q.push(j);
}
}
}
return ans.size() == numCourses ? ans : vector<int>();
}
};func findOrder(numCourses int, prerequisites [][]int) []int {
g := make([][]int, numCourses)
indeg := make([]int, numCourses)
for _, p := range prerequisites {
a, b := p[0], p[1]
g[b] = append(g[b], a)
indeg[a]++
}
q := []int{}
for i, x := range indeg {
if x == 0 {
q = append(q, i)
}
}
ans := []int{}
for len(q) > 0 {
i := q[0]
q = q[1:]
ans = append(ans, i)
for _, j := range g[i] {
indeg[j]--
if indeg[j] == 0 {
q = append(q, j)
}
}
}
if len(ans) == numCourses {
return ans
}
return []int{}
}function findOrder(numCourses: number, prerequisites: number[][]): number[] {
const g: number[][] = Array.from({ length: numCourses }, () => []);
const indeg: number[] = new Array(numCourses).fill(0);
for (const [a, b] of prerequisites) {
g[b].push(a);
indeg[a]++;
}
const q: number[] = [];
for (let i = 0; i < numCourses; ++i) {
if (indeg[i] === 0) {
q.push(i);
}
}
const ans: number[] = [];
while (q.length) {
const i = q.shift()!;
ans.push(i);
for (const j of g[i]) {
if (--indeg[j] === 0) {
q.push(j);
}
}
}
return ans.length === numCourses ? ans : [];
}impl Solution {
pub fn find_order(num_courses: i32, prerequisites: Vec<Vec<i32>>) -> Vec<i32> {
let n = num_courses as usize;
let mut adjacency = vec![vec![]; n];
let mut entry = vec![0; n];
// init
for iter in prerequisites.iter() {
let (a, b) = (iter[0], iter[1]);
adjacency[b as usize].push(a);
entry[a as usize] += 1;
}
// construct deque & reslut
let mut deque = std::collections::VecDeque::new();
for index in 0..n {
if entry[index] == 0 {
deque.push_back(index);
}
}
let mut result = vec![];
// bfs
while !deque.is_empty() {
let head = deque.pop_front().unwrap();
result.push(head as i32);
// update degree of entry
for &out_entry in adjacency[head].iter() {
entry[out_entry as usize] -= 1;
if entry[out_entry as usize] == 0 {
deque.push_back(out_entry as usize);
}
}
}
if result.len() == n {
result
} else {
vec![]
}
}
}public class Solution {
public int[] FindOrder(int numCourses, int[][] prerequisites) {
var g = new List<int>[numCourses];
for (int i = 0; i < numCourses; ++i) {
g[i] = new List<int>();
}
var indeg = new int[numCourses];
foreach (var p in prerequisites) {
int a = p[0], b = p[1];
g[b].Add(a);
++indeg[a];
}
var q = new Queue<int>();
for (int i = 0; i < numCourses; ++i) {
if (indeg[i] == 0) {
q.Enqueue(i);
}
}
var ans = new int[numCourses];
var cnt = 0;
while (q.Count > 0) {
int i = q.Dequeue();
ans[cnt++] = i;
foreach (int j in g[i]) {
if (--indeg[j] == 0) {
q.Enqueue(j);
}
}
}
return cnt == numCourses ? ans : new int[0];
}
}