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2
3

输入：numCourses = 2, prerequisites = [[1,0]]
输出：[0,1]
解释：总共有 2 门课程。要学习课程 1，你需要先完成课程 0。因此，正确的课程顺序为 [0,1] 。

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2
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4

输入：numCourses = 4, prerequisites = [[1,0],[2,0],[3,1],[3,2]]
输出：[0,2,1,3]
解释：总共有 4 门课程。要学习课程 3，你应该先完成课程 1 和课程 2。并且课程 1 和课程 2 都应该排在课程 0 之后。
因此，一个正确的课程顺序是 [0,1,2,3] 。另一个正确的排序是 [0,2,1,3] 。

1
2

输入：numCourses = 1, prerequisites = []
输出：[0]

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struct Solution;
// @lc code=start
impl Solution {
    /// ## 解题思路
    /// - bfs(同207.课程表)
    pub fn find_order(num_courses: i32, prerequisites: Vec<Vec<i32>>) -> Vec<i32> {
        use std::collections::VecDeque;
        let mut in_degrees = vec![0; num_courses as usize];
        let mut g = vec![vec![]; num_courses as usize];
        for p in &prerequisites {
            let (to, from) = (p[0] as usize, p[1] as usize);
            in_degrees[to] += 1;
            g[from].push(to); //
        }

        let mut can_learn = in_degrees
            .iter()
            .enumerate()
            .filter(|(_, &d)| d == 0)
            .map(|(from, _)| from)
            .collect::<VecDeque<_>>();

        let mut res = vec![];
        let mut remain = num_courses as usize;
        while !can_learn.is_empty() {
            remain -= can_learn.len();
            res.append(
                &mut can_learn
                    .iter()
                    .clone()
                    .map(|&i| i as i32)
                    .collect::<Vec<_>>(),
            );
            for _ in 0..can_learn.len() {
                let from = can_learn.pop_front().unwrap();
                for &t in &g[from] {
                    in_degrees[t] -= 1;
                    if in_degrees[t] == 0 {
                        can_learn.push_back(t);
                    }
                }
            }
        }

        if remain == 0 {
            res
        } else {
            vec![]
        }
    }
}
// @lc code=end