Leetcode: 207. 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?

Example 1:

Input: 2, [[1,0]] 
Output: true
Explanation: There are a total of 2 courses to take. 
             To take course 1 you should have finished course 0. So it is possible.

Example 2:

Input: 2, [[1,0],[0,1]]
Output: false
Explanation: 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:

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.

解法：

根据题意，进修一门课之前需要先进修另一门课。这些关系通过一个二维数组作为入参传入函数。其实就是在判断一个有向图是否存在环。为了判断出是否存在环，需要记录本次遍历路径上所有经过的节点。如果遇到已经经过的，那么说明存在环。反之，则不存在环。为了提高效率，我们还可以将从每个节点开始遍历的结果存入一个数组或者map。这样在路径遍历过程中如果发现某个节点之前已经被遍历过，那么直接复用结果就行了。不用再往下重复遍历了。

// Copy from: https://leetcode.com/problems/course-schedule/discuss/480820/Simple-Java-DFS	
public static boolean canFinish(int numCourses, int[][] prerequisites) {
		List[] graph = new ArrayList[numCourses];
		for (int i = 0; i < numCourses; i++) {
			graph[i] = new ArrayList<>();
		}
		
		for (int i = 0; i < prerequisites.length; i++) {
			graph[prerequisites[i][1]].add(prerequisites[i][0]);
		}
		
		boolean[] restoredStack = new boolean[numCourses]; // 暂存路径节点
		boolean[] visited = new boolean[numCourses]; //记录以节点i为起始节点的遍历结果
		for (int i = 0; i < numCourses; i++) {
			if (visited[i])
				continue;
			if (isCycle(visited, restoredStack, i, graph))
				return false;
		}
		return true;
	}

	private static boolean isCycle(boolean[] visited, boolean[] restoredStack, int idx, List[] graph) {
		if (restoredStack[idx])
			return true;
		if (visited[idx])
			return false;
		visited[idx] = true;
		restoredStack[idx] = true;
		for (int i = 0; i < graph[idx].size(); i++) {
			if (isCycle(visited, restoredStack, graph[idx].get(i), graph))
				return true;
		}
		restoredStack[idx] = false;
		return false;
	}