4 BFS & Topological Algorithm

615. Course Schedule

https://www.lintcode.com/problem/course-schedule/description?_from=ladder&&fromId=1

为什么不可以用  for(int[] prerequisite: prerequisites) 呢 （使用Leetcode上的方法）？因为当 prerequisites 的 length 很大的时候，会超时。所以要定义一个edges[]变量，用来记录每一门课程的后续课程。

要注意数据类型的转换：

int pointer = (int)edges[curr].get(i);
indegree[pointer]--;

public class Solution {
    /*
     * @param numCourses: a total of n courses
     * @param prerequisites: a list of prerequisite pairs
     * @return: true if can finish all courses or false
     */
    public boolean canFinish(int numCourses, int[][] prerequisites) {
        // write your code here
        if(prerequisites == null || prerequisites.length == 0 || prerequisites[0].length == 0) {
            return true;
        }
        List[] edges = new List[numCourses];
        int[] indegree = new int[numCourses];
        int count = 0;
        Queue<Integer> queue = new LinkedList<>();
        for(int i = 0; i < numCourses; i++) {
            edges[i] = new LinkedList<>();
        }
        int len = prerequisites.length;
        for(int i = 0; i < len; i++) {
            indegree[prerequisites[i][0]]++;
            edges[prerequisites[i][1]].add(prerequisites[i][0]);
        }
        for(int i = 0; i < numCourses; i++) {
            if(indegree[i] == 0) {
                queue.offer(i);
            }
        }
        while(!queue.isEmpty()) {
            int curr = queue.poll();
            count++;
            int n = edges[curr].size();
            for(int i = 0; i < n; i++) {
                int pointer = (int)edges[curr].get(i);
                indegree[pointer]--;
                if(indegree[pointer] == 0) {
                    queue.offer(pointer);
                }
            }
        }
        return count == numCourses;
    }
}

616. Course Schedule II

https://www.lintcode.com/problem/course-schedule-ii/description?_from=ladder&&fromId=1

public class Solution {
    /*
     * @param numCourses: a total of n courses
     * @param prerequisites: a list of prerequisite pairs
     * @return: the course order
     */
    public int[] findOrder(int numCourses, int[][] prerequisites) {
        // write your code here
        int[] result = new int[numCourses];
        List[] edges = new List[numCourses];
        int[] indegree = new int[numCourses];
        int count = 0;
        Queue<Integer> queue = new LinkedList<>();
        for(int i = 0; i < numCourses; i++) {
            edges[i] = new LinkedList<>();
        }
        int len = prerequisites.length;
        for(int i = 0; i < len; i++) {
            indegree[prerequisites[i][0]]++;
            edges[prerequisites[i][1]].add(prerequisites[i][0]);
        }
        for(int i = 0; i < numCourses; i++) {
            if(indegree[i] == 0) {
                queue.offer(i);
            }
        }
        while(!queue.isEmpty()) {
            int curr = queue.poll();
            result[count] = curr;
            count++;
            int n = edges[curr].size();
            for(int i = 0; i < n; i++) {
                int pointer = (int)edges[curr].get(i);
                indegree[pointer]--;
                if(indegree[pointer] == 0) {
                    queue.offer(pointer);
                }
            }
        }
        if(count != numCourses) {
            return new int[0];
        }
        return result;
    }
}