210. Course Schedule II

Medium

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 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 <= ai, bi < numCourses

• ai != bi

• All the pairs [ai, bi] are distinct.

解題

Runtime: 7 ms, faster than 99.22%

Memory Usage: 6.1 MB, less than 96.50%

func findOrder(numCourses int, prerequisites [][]int) []int {
    // Topological sort
    edges := make([][]int, numCourses)
    indegree := make([]int, numCourses ) //入邊, 入邊 = 0 = 沒有需要先修的課

    for _, pre := range prerequisites {
        edges[pre[1]] = append(edges[pre[1]], pre[0])
        indegree[pre[0]]++
    }

    queue, res := []int{}, []int{}
    for i, v := range indegree { if v == 0 { queue = append(queue, i) } } // 先將沒有入邊的放到 queue   
    for len(queue) != 0 {
        head := queue[0]
        queue = queue[1:]

        res = append(res, head)
        for _, v := range edges[head] {
            indegree[v]-- // 刪除head指向的節點一條入邊 //第head堂課已經上完了，沒有別的先修的話 indegree[v] 會是0
            if indegree[v] == 0 {
                queue = append(queue, v) // 沒有先修課了，可以上這堂課了，加入queue
            }
        }
    }

    if len(res) != numCourses { // 有環，有課沒有上到
        return []int{}
    }

    return res
}