Course Schedule I& II leetcode

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.

思路

先修课程是拓扑排序的经典应用, 这里至关于找有向图是否有环, 若是有环的话拓扑排序能遍历到的节点将少于图的节点. 这里咱们创建一个图, 用一个数组记录每一个节点的入度. 对图进行拓扑排序

复杂度

时间O(V+E) 空间 O(V)

代码

public boolean canFinish(int numCourses, int[][] prerequisites) {
    if (prerequisites == null || prerequisites.length == 0 || prerequisites[0].length == 0) {
        return true;
    }
    //记录入度
    int[] indgree = new int[numCourses];
    //记录有向图的指向节点
    ArrayList[] graph = new ArrayList[numCourses];
    for (int i = 0; i < numCourses; i++) {
        graph[i] = new ArrayList<Integer>();
    }
    //写入有向图的next节点
    for (int i = 0; i < prerequisites.length; i++) {
        graph[prerequisites[i][1]].add(prerequisites[i][0]);
        indgree[prerequisites[i][0]]++;
    }
    Queue<Integer> queue = new LinkedList<Integer>();
    for(int i = 0; i < indgree.length; i++){
        if(indgree[i] == 0){
            queue.add(i);
        }
    }
    int count = 0;
    while (!queue.isEmpty()) {
        int cur = queue.poll();
        count++;
        ArrayList<Integer> list = graph[cur];
        for (Integer tem : list) {
            indgree[tem]--;
            if (indgree[tem] == 0) {
                queue.offer(tem);
            }
        }
    }
    return count == numCourses;
}

Course Schedule II

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, return the ordering of courses you should take to finish all

courses.

There may be multiple correct orders, you just need to return one of them. If it is impossible to finish all courses, return an empty

array.

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 the correct course order is [0,1]

4, [[1,0],[2,0],[3,1],[3,2]]
There are a total of 4 courses to take. To take course 3 you should have finished both courses 1 and 2. Both courses 1 and 2 should

be taken after you finished course 0. So one correct course order is
[0,1,2,3]. Another correct ordering is[0,2,1,3].

思路

与I相同 只是要打印出结果

复杂度

时间O(V+E) 空间 O(V)

代码

public int[] findOrder(int numCourses, int[][] prerequisites) {
        int[] res = new int[numCourses];
        //记录入度
        int[] indgree = new int[numCourses];
        //记录有向图的指向节点
        ArrayList[] graph = new ArrayList[numCourses];
        for (int i = 0; i < numCourses; i++) {
            graph[i] = new ArrayList<Integer>();
        }
        //写入有向图的next节点
        for (int i = 0; i < prerequisites.length; i++) {
            graph[prerequisites[i][1]].add(prerequisites[i][0]);
            indgree[prerequisites[i][0]]++;
        }
        Queue<Integer> queue = new LinkedList<Integer>();
        for(int i = 0; i < indgree.length; i++){
            if(indgree[i] == 0){
                queue.add(i);
            }
        }
        int count = 0;
        while (!queue.isEmpty()) {
            int cur = queue.poll();
            res[count++] = cur;
            ArrayList<Integer> list = graph[cur];
            for (Integer tem : list) {
                indgree[tem]--;
                if (indgree[tem] == 0) {
                    queue.offer(tem);
                }
            }
        }
        return count != numCourses ? new int[0] : res;
    }