207. Course Schedule

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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:

1
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.

1
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.

解法1：有向图

这题和Course Schedule II的解法基本一样，就是判断有向图中是否有环。基本的入度出度的解法要掌握。
C++

1

Java

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public class Solution {
    public boolean canFinish(int numCourses, int[][] prerequisites) {
        if (prerequisites == null || prerequisites.length == 0) {
            return true;
        }
        // no circle in the graph + number of visited nodes == numCourses
        int[] ins = new int[numCourses];
        Map<Integer, List<Integer>> map = new HashMap<Integer, List<Integer>>();
        for (int[] pre : prerequisites) {
            if (!map.containsKey(pre[1])) {
                map.put(pre[1], new ArrayList<Integer>());
            }
            map.get(pre[1]).add(pre[0]);    // because of no duplicate edges in the input prerequisities
            ins[pre[0]]++;  // add the in-count for the node
        }
        Queue<Integer> queue = new LinkedList<Integer>();
        for (int i = 0; i < numCourses; i++) {
            if (ins[i] == 0) {
                queue.offer(i);
            }
        }
        while (!queue.isEmpty()) {
            int root = queue.poll();
            if (map.containsKey(root)) {
                List<Integer> children = map.get(root);
                for (int child : children) {
                    ins[child] -= 1;
                    if (ins[child] == 0) {
                        queue.offer(child);
                    }
                }
            }
        }
        // check if there is non-zero ins
        for (int i = 0; i < numCourses; i++) {
            if (ins[i] != 0) {
                return false;
            }
        }
        return true;
    }
}