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?

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.
You may assume that there are no duplicate edges in the input prerequisites.

click to show more hints.

给出课程的门数numCourses和各个课程的先修要求，那么可以抽象出一张图（Graphy），每门课程作为一个节点（node），先修要求作为有向边（direct-edge），实际上本题是求：给出的这幅有向图中，是否存在环路？这个问题在算法导论中的拓扑排序中介绍过：有向图中如果不存在环，那么在深度搜索（DFS）的过程中不存在后向边。

那么可以直接对每个节点依次进行DFS，维护一个visited数组，来记录是否访问过节点i，这里需要注意的是，一个节点可能有多个祖先节点，我们可以学习算法导论中的DFS方法，对节点进行染色：1）没有访问过的节点染成白色；2）入栈后的节点染成灰色；3）访问后出栈的节点染成黑色。这样做的好处是，染成灰色的节点是我们当前遍历的路径下的节点，染成黑色的节点则是之前遍历过的路径节点。已经遍历过、并且不存在环路的路径节点被染成黑色，那么下次遍历到这个节点时，无需向下遍历，肯定不存在环路；如果遍历的时候访问到了灰色节点，那么说明当前的路径存在一条后向边（环路），那么直接返回false即可。

class Solution {
private int[] visited;
private Map<Integer, List<Integer>> map;

public boolean canFinish(int numCourses, int[][] prerequisites) {
if (numCourses <= 1 || prerequisites == null || prerequisites.length == 0)
return true;

visited = new int[numCourses];
map = new HashMap<>();

for (int[] pre: prerequisites) {
if (map.containsKey(pre[0])) {
map.get(pre[0]).add(pre[1]);
}
else {
List<Integer> list = new ArrayList<Integer>();
list.add(pre[1]);
map.put(pre[0], list);
}
}

for (int i = 0; i < numCourses; ++i) {
if (map.containsKey(i) && visited[i] == 0) {
if (!visit(map, i))
return false;
}
}

return true;
}

private boolean visit(Map<Integer, List<Integer>> map, int i) {
if (visited[i] == 1)
return false;
else if (visited[i] == 2 || !map.containsKey(i))
return true;

boolean res = true;

visited[i] = 1;
for (Integer next : map.get(i)) {
if (!visit(map, next)) {
res = false;
break;
}
}
visited[i] = 2;

return res;
}

}