Course Schedule I II LeetCode Java

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.

这道题目考的是拓扑排序相关内容，不了解什么是拓扑排序，可以参考博客：http://blog.csdn.net/dm_vincent/article/details/7714519

所以，这道题目也有两种解法(程序参考自：https://leetcode.com/discuss/39456/java-dfs-and-bfs-solution)：

BFS：

public class Solution {    public boolean canFinish(int numCourses, int[][] prerequisites) {        ArrayList[] graph = new ArrayList[numCourses];        int[] degree = new int[numCourses];        Queue queue = new LinkedList();        int count=0;        for(int i=0;i<numCourses;i++)            graph[i] = new ArrayList();        for(int i=0; i<prerequisites.length;i++){            degree[prerequisites[i][1]]++;            graph[prerequisites[i][0]].add(prerequisites[i][1]);        }        for(int i=0; i<degree.length;i++){            if(degree[i] == 0){                queue.add(i);                count++;            }        }        while(queue.size() != 0){            int course = (int)queue.poll();            for(int i=0; i<graph[course].size();i++){                int pointer = (int)graph[course].get(i);                degree[pointer]--;                if(degree[pointer] == 0){                    queue.add(pointer);                    count++;                }            }        }        if(count == numCourses)            return true;        else                return false;    }}

DFS:

        public boolean canFinishDFS(int numCourses, int[][] prerequisites) {            ArrayList[] graph = new ArrayList[numCourses];            for(int i=0;i<numCourses;i++)                graph[i] = new ArrayList();            boolean[] visited = new boolean[numCourses];            boolean[] visiting=new boolean[numCourses];            for(int i=0; i<prerequisites.length;i++){                graph[prerequisites[i][1]].add(prerequisites[i][0]);            }            for(int i=0; i<numCourses; i++){                if(!dfs(graph,visited,visiting,i))                    return false;            }            return true;        }        private boolean dfs(ArrayList[] graph, boolean[] visited, boolean[] visiting, int course){            if(visiting[course])                return false;            if(visited[course])                return true;            visiting[course] = true;;            for(int i=0; i<graph[course].size();i++){                if(!dfs(graph,visited,visiting,(int)graph[course].get(i)))                    return false;            }            visiting[course]=false;            visited[course] = true;            return true;        }

Course Schedule II

这道题目和上道题目思路类似，只不过要求输出选课的顺序

同样有两种解法

BFS：

    public int[] findOrder(int numCourses, int[][] prerequisites){        ArrayList<Integer>[] graph=new ArrayList[numCourses];        int degree[] =new int[numCourses];        Queue<Integer> queue=new LinkedList<Integer>();        int count=0;        int[] rst=new int[numCourses];        for(int i=0;i<numCourses;i++){            graph[i]=new ArrayList<Integer>();        }        for(int i=0;i<prerequisites.length;i++){            degree[prerequisites[i][0]]++;            graph[prerequisites[i][1]].add(prerequisites[i][0]);        }        for(int i=0;i<degree.length;i++){            if(degree[i]==0){                queue.add(i);                rst[count]=i;                count++;            }        }        while(queue.size()!=0){            int course=(int)queue.poll();            for(int i=0;i<graph[course].size();i++){                int pointer=(int)graph[course].get(i);                degree[pointer]--;                if(degree[pointer]==0){                    queue.add(pointer);                    rst[count]=pointer;                    count++;                }            }        }        if(count == numCourses)            return rst;        else                return new int[0];    } 

DFS：

public class Solution {        private int[] rst;        private int count;        public int[] findOrder(int numCourses, int[][] prerequisites) {            ArrayList[] graph = new ArrayList[numCourses];            rst=new int[numCourses];            count=0;            for(int i=0;i<numCourses;i++)                graph[i] = new ArrayList();            boolean[] visited = new boolean[numCourses];            boolean[] visiting=new boolean[numCourses];            for(int i=0; i<prerequisites.length;i++){                graph[prerequisites[i][0]].add(prerequisites[i][1]);            }            for(int i=0; i<numCourses; i++){                if(!dfs2(graph,visited,visiting,i))                    return new int[0];            }            return rst;        }        private boolean dfs2(ArrayList[] graph, boolean[] visited, boolean[] visiting, int course){            if(visiting[course])                return false;            if(visited[course])                return true;            visiting[course] = true;;            for(int i=0; i<graph[course].size();i++){                if(!dfs2(graph,visited,visiting,(int)graph[course].get(i)))                    return false;            }            visiting[course]=false;            visited[course] = true;            rst[count]=course;            count++;            return true;        }}