Course Schedule ——LeetCode

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

 

题目大意：给一堆课程依赖，找出是否可以顺利修完全部课程，拓扑排序。

解法一：DFS+剪枝，DFS探测是否有环，图的表示采用矩阵。

    public boolean canFinish(int n, int[][] p) {
        if (p == null || p.length == 0) {
            return true;
        }
        int row = p.length;
        int[][] pre = new int[n][n];
        for (int i = 0; i < n; i++) {
            Arrays.fill(pre[i], -1);
        }
        for (int i = 0; i < row; i++) {
            pre[p[i][0]][p[i][1]] = p[i][1];
        }
//        System.out.println(Arrays.deepToString(pre));
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (i == j) {
                    continue;
                }
                if (pre[i][j] != -1) {
                    Deque<Integer> queue = new ArrayDeque<>();
                    queue.offer(pre[i][j]);
                    Set<Integer> circleDep = new HashSet<>();
                    circleDep.add(pre[i][j]);
                    while (!queue.isEmpty()) {
                        int dep = queue.poll();
                        if (dep >= row) {
                            continue;
                        }
                        for (int k = 0; k < n; k++) {
                            if (pre[dep][k] == -1) {
                                continue;
                            }
                            if (circleDep.contains(pre[dep][k])) {
                                return false;
                            }
                            queue.offer(pre[dep][k]);
                            circleDep.add(pre[dep][k]);
                            pre[dep][k]=-1;
                        }
                    }
                }
            }
        }
        return true;
    }

 

解法二：BFS，参考别人的思路，也是用矩阵表示图，另外用indegree表示入度，先把入度为0的加入队列，当队列非空，逐个取出队列中的元素，indegree[i]-1==0的继续入队列，BFS遍历整个图，用count记录课程数，如果等于给定值则返回true。

    public boolean canFinish(int n, int[][] p) {
        if (p == null || p.length == 0) {
            return true;
        }
        int[][] dep = new int[n][n];
        int[] indegree = new int[n];
        for(int i=0;i<p.length;i++){
            if(dep[p[i][0]][p[i][1]]==1){
                continue;
            }
            dep[p[i][0]][p[i][1]]=1;
            indegree[p[i][1]]++;
        }
        Deque<Integer> queue = new ArrayDeque<>();
        for(int i=0;i<n;i++){
            if(indegree[i]==0){
                queue.offer(i);
            }
        }
        int count = 0;
        while(!queue.isEmpty()){
            count++;
            int cos = queue.poll();
            for(int i=0;i<n;i++){
                if(dep[cos][i]!=0){
                    if(--indegree[i]==0){
                        queue.offer(i);
                    }
                }
            }
        }
        return count==n;
    }