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82 lines
2.8 KiB
Markdown
Executable File
82 lines
2.8 KiB
Markdown
Executable File
# [207. Course Schedule](https://leetcode.com/problems/course-schedule/)
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## 题目
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There are a total of n courses you have to take, labeled from `0` to `n-1`.
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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]`
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Given the total number of courses and a list of prerequisite **pairs**, is it possible for you to finish all courses?
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**Example 1**:
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Input: 2, [[1,0]]
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Output: true
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Explanation: There are a total of 2 courses to take.
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To take course 1 you should have finished course 0. So it is possible.
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**Example 2**:
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Input: 2, [[1,0],[0,1]]
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Output: false
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Explanation: There are a total of 2 courses to take.
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To take course 1 you should have finished course 0, and to take course 0 you should
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also have finished course 1. So it is impossible.
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**Note**:
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1. The input prerequisites is a graph represented by **a list of edges**, not adjacency matrices. Read more about [how a graph is represented](https://www.khanacademy.org/computing/computer-science/algorithms/graph-representation/a/representing-graphs).
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2. You may assume that there are no duplicate edges in the input prerequisites.
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## 题目大意
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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，判断是否可能完成所有课程的学习？
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## 解题思路
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- 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。
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- 这一题就是标准的 AOV 网的拓扑排序问题。拓扑排序问题的解决办法是主要是循环执行以下两步，直到不存在入度为0的顶点为止。
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- 1. 选择一个入度为0的顶点并输出之；
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- 2. 从网中删除此顶点及所有出边。
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循环结束后，若输出的顶点数小于网中的顶点数，则输出“有回路”信息，即无法完成所有任务；否则输出的顶点序列就是一种拓扑序列，即可以完成所有任务。
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## 代码
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```go
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package leetcode
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// AOV 网的拓扑排序
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func canFinish(n int, pre [][]int) bool {
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in := make([]int, n)
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frees := make([][]int, n)
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next := make([]int, 0, n)
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for _, v := range pre {
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in[v[0]]++
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frees[v[1]] = append(frees[v[1]], v[0])
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}
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for i := 0; i < n; i++ {
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if in[i] == 0 {
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next = append(next, i)
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}
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}
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for i := 0; i != len(next); i++ {
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c := next[i]
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v := frees[c]
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for _, vv := range v {
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in[vv]--
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if in[vv] == 0 {
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next = append(next, vv)
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}
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}
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}
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return len(next) == n
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}
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``` |