Course Schedule-优快云博客

本文介绍了一个算法，用于判断在存在先修课程约束的情况下，是否有可能完成所有课程。通过使用图论中的拓扑排序思想，该算法能有效地解决课程依赖问题。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

here 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.

public class Solution {
    public boolean canFinish(int numCourses, int[][] prerequisites) {
        ArrayList<Set<Integer>> arrayList = new ArrayList<>();
        for (int i = 0; i < numCourses; i++) {
        	arrayList.add(new HashSet<Integer>());
		}
        for (int i = 0; i < prerequisites.length; i++) {
        	arrayList.get(prerequisites[i][1]).add(prerequisites[i][0]);
		}
        int[] preNum = new int[numCourses];
        for (int i = 0; i < numCourses; i++) {
        	Set<Integer> set = arrayList.get(i);
        	Iterator<Integer> iterator = set.iterator();
        	while (iterator.hasNext()) {
        		preNum[iterator.next()]++;
        	}
		}
        for (int i = 0; i < numCourses; i++) {
			int j = 0;
			for (; j < numCourses; j++) {
				if (preNum[j] == 0) {
					break;
				}
			}
			if (j == numCourses) {
				return false;
			}
			preNum[j] = -1;
			Set<Integer> set = arrayList.get(j);
			Iterator<Integer> iterator = set.iterator();
			while (iterator.hasNext()) {
        		preNum[iterator.next()]--;
        	}
		}
        return true;
    }
}