LeetCode OJ 207. Course Schedule 拓扑排序+邻接表

时间：2016-06-17 16:41 来源：作者：收藏

转载请标明出处： http://blog.csdn.net/lmj623565791/article/details/47079737；本文出自:【张鸿洋的博客】1、概述话说最近股市变动不变，也成了热火朝天的话题。不知道大家有没有考虑做个实时更新股市数据的app呢？假设我们要做一个股市数据实时更新的app

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

207. Course Schedule

My SubmissionsTotal Accepted: 42752 Total Submissions: 154752 Difficulty: Medium

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.

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这是一道图论的题目，考察的是拓扑排序。拓扑排序需要邻接表存储图。四年没用邻接表了，复习了下邻接表再来做的这个题目。

拓扑排序的思路：

入度为0的顶点入队列
队列每出一个顶点，这个顶点所有边的另一头元素的入度减一。如果入度减到0了，入队列。
循环第二步，直至队列为空。

我的AC代码：

package march;

import java.util.ArrayDeque;
import java.util.Queue;

/**
 * Created by Administrator on 2016/6/17.
 */
public class CourseSchedule {

	public static void main(String[] args) {
		CourseSchedule cs = new CourseSchedule();
		int[][]        a  = {{0, 1}};
		System.out.println(cs.canFinish(2, a));
		int[][] b = {{0, 1}, {1, 0}};
		System.out.println(cs.canFinish(2, b));
	}

	public boolean canFinish(int numCourses, int[][] prerequisites) {
		Edge[]    edges   = new Edge[numCourses];
		int[]     indgree = new int[numCourses];
		boolean[] visited = new boolean[numCourses];
		for (int i = 0; i < numCourses; i++) edges[i] = new Edge(-1);

		for (int i = 0; i < prerequisites.length; i++) {
			int[] edge = prerequisites[i];
			Edge e = new Edge(edge[1]);
			int edgeHead = edge[0];
			e.next = edges[edgeHead].next;
			edges[edgeHead].next = e;

			indgree[edge[1]]++;
		}
		Queue<Integer> q = new ArrayDeque<Integer>(numCourses);
		for (int i = 0; i < numCourses; i++) {
			if (indgree[i] == 0) q.add(i);
		}
		if (q.isEmpty()) return false;

		while (!q.isEmpty()) {
			int ver = q.poll();
			visited[ver] = true;
			Edge head = edges[ver].next;
			while (head != null) {
				if (--indgree[head.num] == 0) q.add(head.num);
				head = head.next;
			}
		}

		boolean result = true;
		for (boolean v : visited) result &= v;
		return result;
	}
}

class Edge {
	public Edge(int n) {
		num = n;
	}

	public int num;
	public Edge next = null;
}