210. Course Schedule II

[Tcdian]

210. Course Schedule II

现在你总共有 n 门课需要选，记为 0 到 n-1。

在选修某些课程之前需要一些先修课程。 例如，想要学习课程 0 ，你需要先完成课程 1 ，我们用一个匹配来表示他们: [0,1]

给定课程总量以及它们的先决条件，返回你为了学完所有课程所安排的学习顺序。

可能会有多个正确的顺序，你只要返回一种就可以了。如果不可能完成所有课程，返回一个空数组。

Example 1

Input: 2, [[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: 4, [[1,0],[2,0],[3,1],[3,2]]
Output: [0,1,2,3] or [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] .

Note

• 输入的先决条件是由边缘列表表示的图形，而不是邻接矩阵。
• 你可以假定输入的先决条件中没有重复的边。

[Tcdian]

Solution

• JavaScript Solution

/**
 * @param {number} numCourses
 * @param {number[][]} prerequisites
 * @return {number[]}
 */
var findOrder = function(numCourses, prerequisites) {
    const graph = Array.from(new Array(numCourses), () => []);
    for (let i = 0; i < prerequisites.length; i++) {
        const [course, prerequisite] = prerequisites[i];
        graph[prerequisite].push(course);
    }
    const Status = {
        Initial: 0,
        Visiting: 1,
        Visited: 2,
    };
    const statuses = new Array(numCourses).fill(Status.Initial);
    const result = [];
    for (let i = 0; i < numCourses; i++) {
        if (dfsCourses(i)) {
            return [];
        }
    }
    return result.reverse();

    function dfsCourses(course) {
        if (statuses[course] === Status.Visiting) {
            return true;
        }
        if (statuses[course] === Status.Visited) {
            return false;
        }
        statuses[course] = Status.Visiting;
        const advancedCourse = graph[course];
        for (let i = 0; i < advancedCourse.length; i++) {
            if (dfsCourses(advancedCourse[i])) {
                return true;
            }
        }
        statuses[course] = Status.Visited;
        result.push(course);
        return false;
    }
};

• TypeScript Solution

enum Status {
    Initial,
    Visiting,
    Visited,
}

function findOrder(numCourses: number, prerequisites: number[][]): number[] {
    const graph: number[][] = Array.from(new Array(numCourses), () => []);
    for (let i = 0; i < prerequisites.length; i++) {
        const [course, prerequisite] = prerequisites[i];
        graph[prerequisite].push(course);
    }
    const statuses: Status[] = new Array(numCourses).fill(Status.Initial);
    const result: number[] = [];
    for (let i = 0; i < numCourses; i++) {
        if (dfsCourses(i)) {
            return [];
        }
    }
    return result.reverse();

    function dfsCourses(course: number): boolean {
        if (statuses[course] === Status.Visiting) {
            return true;
        }
        if (statuses[course] === Status.Visited) {
            return false;
        }
        statuses[course] = Status.Visiting;
        const advancedCourse = graph[course];
        for (let i = 0; i < advancedCourse.length; i++) {
            if (dfsCourses(advancedCourse[i])) {
                return true;
            }
        }
        statuses[course] = Status.Visited;
        result.push(course);
        return false;
    }
};