LeetCode 207. Course Schedule（课程安排）

原题网址：https://leetcode.com/problems/course-schedule/

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, is it possible for you to finish all courses?

For example:

2, [[1,0]]

There are a total of 2 courses to take. To take course 1 you should have finished course 0. So it is possible.

2, [[1,0],[0,1]]

There are a total of 2 courses to take. To take course 1 you should have finished course 0, and to take course 0 you should also have finished course 1. So it is impossible.

Note:

The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how
a graph is represented.

click to show more hints.

Hints:

This problem is equivalent to finding if a cycle exists in a directed graph. If a cycle exists, no topological ordering exists and therefore it will be impossible to take all courses.
Topological Sort via DFS - A great video tutorial (21 minutes) on Coursera explaining
the basic concepts of Topological Sort.
Topological sort could also be done via BFS.

方法一：广度优先搜索。

public class Solution {
public boolean canFinish(int numCourses, int[][] prerequisites) {
int[] pre = new int[numCourses];
List<Integer>[] satisfies = new List[numCourses];
for(int i=0; i<numCourses; i++) satisfies[i] = new ArrayList<>();
for(int i=0; i<prerequisites.length; i++) {
satisfies[prerequisites[i][1]].add(prerequisites[i][0]);
pre[prerequisites[i][0]] ++;
}
int finish = 0;
LinkedList<Integer> queue = new LinkedList<>();
for(int i=0; i<numCourses; i++) {
if (pre[i] == 0) queue.add(i);
}
while (!queue.isEmpty()) {
int course = queue.remove();
finish ++;
if (satisfies[course] == null) continue;
for(int c: satisfies[course]) {
pre[c] --;
if (pre[c] == 0) queue.add(c);
}
}
return finish == numCourses;
}
}

方法二：深度优先搜索。

public class Solution {
private boolean[] canFinish;
private boolean[] visited;
private List<Integer>[] depends;
private boolean canFinish(int course) {
if (visited[course]) return canFinish[course];
visited[course] = true;
for(int c: depends[course]) {
if (!canFinish(c)) return false;
}
canFinish[course] = true;
return canFinish[course];
}
public boolean canFinish(int numCourses, int[][] prerequisites) {
canFinish = new boolean[numCourses];
visited = new boolean[numCourses];
depends = new List[numCourses];
for(int i=0; i<numCourses; i++) depends[i] = new ArrayList<Integer>();
for(int i=0; i<prerequisites.length; i++) {
depends[prerequisites[i][0]].add(prerequisites[i][1]);
}
for(int i=0; i<numCourses; i++) {
if (!canFinish(i)) return false;
}
return true;
}
}