【LeetCode】Course Schedule 解题报告

【题目】

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.

【解析】

典型的拓扑排序。原理也很简单，在一个有向图中，每次找到一个没有前驱节点的节点（也就是入度为0的节点），然后把它指向其他节点的边都去掉，重复这个过程（BFS），直到所有节点已被找到，或者没有符合条件的节点（如果图中有环存在）。

回顾一下图的三种表示方式：边表示法（即题目中表示方法），邻接表法，邻接矩阵。用邻接表存储图比较方便寻找入度为0的节点。

【Java代码】

public class Solution {
    public boolean canFinish(int numCourses, int[][] prerequisites) {
        // init the adjacency list
        List<Set> posts = new ArrayList<Set>();
        for (int i = 0; i < numCourses; i++) {
            posts.add(new HashSet<Integer>());
        }

        // fill the adjacency list
        for (int i = 0; i < prerequisites.length; i++) {
            posts.get(prerequisites[i][1]).add(prerequisites[i][0]);
        }

        // count the pre-courses
        int[] preNums = new int[numCourses];
        for (int i = 0; i < numCourses; i++) {
            Set set = posts.get(i);
            Iterator<Integer> it = set.iterator();
            while (it.hasNext()) {
            	preNums[it.next()]++;
            }
        }

        // remove a non-pre course each time
        for (int i = 0; i < numCourses; i++) {
            // find a non-pre course
            int j = 0;
            for ( ; j < numCourses; j++) {
                if (preNums[j] == 0) break;
            }

            // if not find a non-pre course
            if (j == numCourses) return false;

            preNums[j] = -1;

            // decrease courses that post the course
            Set set = posts.get(j);
            Iterator<Integer> it = set.iterator();
            while (it.hasNext()) {
            	preNums[it.next()]--;
            }
        }

        return true;
    }
}

注意，输入可能有重复的边，所以邻接表用HashSet存储。

下面一种代码是不用HashSet的，对于重复的边，它在邻接表中村了两份，同时计算入度时也算了两次，所以代码不会有问题。但个人感觉最好用HashSet，这样符合图的定义。

下面的代码还是比较典型的BFS写法，大家可以对比理解下：

public class Solution {
    public boolean canFinish(int numCourses, int[][] prerequisites) {
        List<List<Integer>> posts = new ArrayList<List<Integer>>();
        for (int i = 0; i < numCourses; i++) {
            posts.add(new ArrayList<Integer>());
        }

        int[] preNums = new int[numCourses];
        for (int i = 0; i < prerequisites.length; i++) {
            posts.get(prerequisites[i][1]).add(prerequisites[i][0]);
            preNums[prerequisites[i][0]]++;
        }

        Queue<Integer> queue = new LinkedList<Integer>();
        for (int i = 0; i < numCourses; i++) {
            if (preNums[i] == 0){
                queue.offer(i);
            }
        }

        int count = numCourses;
        while (!queue.isEmpty()) {
            int cur = queue.poll();
            for (int i : posts.get(cur)) {
                if (--preNums[i] == 0) {
                    queue.offer(i);
                }
            }
            count--;
        }

        return count == 0;
    }
}