[LeetCode#210]Course Schedule II

Problem:

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].

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

 

Analysis:

This problem is an updated version of "Course Schedule". The only extra thing we need to do is to use a array to record the order those courses were unlocked. Once a element was poped out of the queue, we add into our result array.
        while (!queue.isEmpty()) {
            int cur = queue.poll();
            visited[cur] = true;
            ret[count] = cur;
            count++;
            ...
        }
But don't be happy too earily!!! For course schedule one we have written
int len = prerequisites.length;
if (len == 0)
    return true;

For this problem, I have mistakely update it into:
int len = prerequisites.length;
if (len == 0)
    return new int[0];

Which causes the error:
Input:
1, []
Output:
[]
Expected:
Special judge: No expected output available.

Actually, all numCourses are inherently unlocked, and the programm logic could handle this case:
for (int i = 0; i < numCourses; i++) {
    //all pre counter is 0
    if (pre_counter[i] == 0) {
        queue.offer(i);
    }
}
Then
while (!queue.isEmpty()) {
    int cur = queue.poll();
    visited[cur] = true;
    ret[count] = cur;
    ...
}

 

Solution:

public class Solution {
    public int[] findOrder(int numCourses, int[][] prerequisites) {
        if (prerequisites == null)
            throw new IllegalArgumentException("the prerequisites matrix is not valid");
        int len = prerequisites.length;
        boolean[] visited = new boolean[numCourses];
        int[] ret = new int[numCourses];
        int[] pre_counter = new int[numCourses];
        int count = 0;
        Queue<Integer> queue = new LinkedList<Integer> ();
        for (int i = 0; i < len; i++) {
            pre_counter[prerequisites[i][0]]++;
        }
        for (int i = 0; i < numCourses; i++) {
            if (pre_counter[i] == 0) {
                queue.offer(i);
            }
        }
        while (!queue.isEmpty()) {
            int cur = queue.poll();
            visited[cur] = true;
            ret[count] = cur;
            count++;
            for (int i = 0; i < len; i++) {
                if (prerequisites[i][1] == cur) {
                    pre_counter[prerequisites[i][0]]--;
                    if (pre_counter[prerequisites[i][0]] == 0 && visited[prerequisites[i][0]] == false)
                        queue.offer(prerequisites[i][0]);
                }
            }
        }
        if (count == numCourses)
            return ret;
        else
            return new int[0];
    }
}

 

转载于:https://www.cnblogs.com/airwindow/p/4765370.html