210. Course Schedule II #

题目 #

There are a total of n courses you have to take, labeled from 0 to n-1.

Some courses may have prerequisites, for example to take course 0 you have to first take course 1, which is expressed as a pair: [0,1]

Given the total number of courses and a list of prerequisite pairs, return the ordering of courses you should take to finish all courses.

There may be multiple correct orders, you just need to return one of them. If it is impossible to finish all courses, return an empty array.

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:

1. The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about  how a graph is represented.
2. You may assume that there are no duplicate edges in the input prerequisites.

题目大意 #

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。

解题思路 #

• 给出 n 个任务，每两个任务之间有相互依赖关系，比如 A 任务一定要在 B 任务之前完成才行。问是否可以完成所有任务，如果可以完成任务，就输出完成任务的顺序，如果不能完成，输出空数组。
• 这一题是第 207 题的加强版。解题思路是 AOV 网的拓扑排序。最后输出数组即可。代码和第 207 题基本不变。具体解题思路见第 207 题。

代码 #

package leetcode

func findOrder(numCourses int, prerequisites [][]int) []int {
	in := make([]int, numCourses)
	frees := make([][]int, numCourses)
	next := make([]int, 0, numCourses)
	for _, v := range prerequisites {
		in[v[0]]++
		frees[v[1]] = append(frees[v[1]], v[0])
	}
	for i := 0; i < numCourses; i++ {
		if in[i] == 0 {
			next = append(next, i)
		}
	}
	for i := 0; i != len(next); i++ {
		c := next[i]
		v := frees[c]
		for _, vv := range v {
			in[vv]--
			if in[vv] == 0 {
				next = append(next, vv)
			}
		}
	}
	if len(next) == numCourses {
		return next
	}
	return []int{}
}