leetcode 207. Course Schedule 課程計劃 ---------- java

標籤：oid   sse   ++   remove   expr   number   expressed   logs   cad   

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.

Note:

1. The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.
2. You may assume that there are no duplicate edges in the input prerequisites。
   
   

主要考察的圖的遍曆。

自己寫的都逾時了，說明還是有很多問題。

1、暴力演算法，結果逾時。

public class Solution {    public boolean canFinish(int numCourses, int[][] prerequisites) {                int[][] num = new int[numCourses][numCourses];        for (int i = 0; i < prerequisites.length; i++){            num[prerequisites[i][0]][prerequisites[i][1]] = 1;        }        int[] finish = new int[numCourses];        boolean flag = false;        while (true){            flag = false;            for (int i = 0; i < numCourses; i++){                for (int j = 0; j < numCourses; j++){                    if (num[i][j] == 1){                        break;                    } else if (j == numCourses - 1 && finish[i] == 0){                        finish[i] = 1;                        isfinish(num, i);                        flag = true;                    }                }            }            if (flag == false){                for (int i = 0; i < numCourses; i++){                    if (finish[i] == 0)                        break;                    else if (i == numCourses - 1)                        return true;                }                return false;            }        }    }    public void isfinish(int[][] num, int pos){        for (int i = 0; i < num.length; i++){            num[i][pos] = 0;        }    }}

 

 

2、最佳化一下，然而還是逾時。

public class Solution {    public boolean canFinish(int numCourses, int[][] prerequisites) {                int[][] num = new int[numCourses][numCourses];        for (int i = 0; i < prerequisites.length; i++){            if (num[prerequisites[i][1]][prerequisites[i][0]] == 1){                return false;            }            num[prerequisites[i][0]][prerequisites[i][1]] = 1;            for (int j = 0; j < numCourses; j++){                /*                k = prerequisites[i][1]之前需要完成的工作(也就是num[k][j] == 1)，都需要完成                num[num[prerequisites[i][0]][j] = 1;                */                if (num[prerequisites[i][1]][j] == 1){                    num[prerequisites[i][0]][j] = 1;                }                /*                需要先完成k = prerequisites[i][0]才能完成的工作(num[j][k] == 1)，也要完成 num[j][prerequisites[i][1]] = 1;                */                if (num[j][prerequisites[i][0]] == 1){                    num[j][prerequisites[i][1]] = 1;                }            }                    }        return true;    }    }

 

3、DFS（參考discuss）

public class Solution {    public boolean canFinish(int numCourses, int[][] prerequisites) {                ArrayList[] list = new ArrayList[numCourses];        for (int i = 0; i < numCourses; i++){            list[i] = new ArrayList<Integer>();        }        for (int i = 0; i < prerequisites.length; i++){            list[prerequisites[i][1]].add(prerequisites[i][0]);        }        boolean[] visit = new boolean[numCourses];        for (int i = 0; i < numCourses; i++){            if (!dfs(list, visit, i)){                return false;            }        }        return true;    }        public boolean dfs(ArrayList[] list, boolean[] visit, int pos){        if (visit[pos]){            return false;        } else {            visit[pos] = true;        }        for (int i = 0; i < list[pos].size(); i++){            if (!dfs(list, visit, (int) list[pos].get(i))){                return false;            }
　　　　　　　list[pos].remove(i);        }        visit[pos] = false;        return true;    }    }

 

 

 

4、BFS

public class Solution {    public boolean canFinish(int numCourses, int[][] prerequisites) {        List<Integer>[] adj = new List[numCourses];            for(int i = 0; i < numCourses; i++)            adj[i] = new ArrayList<Integer>();        int[] indegree = new int[numCourses];                  Queue<Integer> readyCourses = new LinkedList();         int finishCount = 0;                                for (int i = 0; i < prerequisites.length; i++)          {            int curCourse = prerequisites[i][0];                    int preCourse = prerequisites[i][1];                    adj[preCourse].add(curCourse);            indegree[curCourse]++;        }        for (int i = 0; i < numCourses; i++)         {            if (indegree[i] == 0)                 readyCourses.offer(i);                   }        while (!readyCourses.isEmpty())         {            int course = readyCourses.poll();        // finish            finishCount++;            for (int nextCourse : adj[course])             {                indegree[nextCourse]--;                if (indegree[nextCourse] == 0)                        readyCourses.offer(nextCourse);  // ready            }        }        return finishCount == numCourses;    }}

 

leetcode 207. Course Schedule 課程計劃 ---------- java