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【Leetcode】Course Schedule

时间:2016-05-23 15:15:28      阅读:234      评论:0      收藏:0      [点我收藏+]

题目链接:
题目:https://leetcode.com/problems/course-schedule/

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.

思路:
拓扑排序,计算无环结点个数,

算法:

    public boolean canFinish(int numCourses, int[][] prerequisites) {
        int preCount[] = new int[numCourses];
        Queue<Integer> notPre = new LinkedList<Integer>();
        // 统计每个结点作为前驱的次数
        for (int i = 0; i < prerequisites.length; i++) {
            preCount[prerequisites[i][0]]++;
        }
        // 记录出度为0的结点
        for (int i = 0; i < preCount.length; i++) {
            if (preCount[i] == 0) {
                notPre.offer(i);
            }
        }
        int num = notPre.size();// 可以拓扑排序的节点个数
        while (!notPre.isEmpty()) {//BFS
            int node = notPre.poll();
            for (int i = 0; i < prerequisites.length; i++) {
                if (prerequisites[i][1] == node) {
                    preCount[prerequisites[i][0]]--;//删除和node有关的边
                    if (preCount[prerequisites[i][0]] == 0) {//如果出度为0
                        num++;
                        notPre.offer(prerequisites[i][0]);
                    }
                }
            }
        }

        return num == numCourses;
    }

【Leetcode】Course Schedule

原文:http://blog.csdn.net/yeqiuzs/article/details/51476915

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