Middle-题目86：207. Course Schedule

题目原文：

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.

题目大意：

你要上n门课，课的编号分别是0,…n-1.

有一些课程有先修课，例如要学习课程0需先学习课程1，这样用一个有序数对[0,1]表示。

给出课程的数目和先修课程的有序数对列表，判断是否能修完所有课程。

题目分析：

这也是一道关于图的问题。用有向图来表示课程的先修关系，例如数对[0.1]对应有向图中0->1的边。判断是否存在拓扑序列即可。

源码：（language：java）

public class Solution {
public  boolean canFinish(int numCourses, int[][] prerequisites) {
Map<Integer, Set<Integer>> adjList = new HashMap<Integer,Set<Integer>>();
int[] indegree = new int[numCourses];
for(int i = 0; i<numCourses;i++)
adjList.put(i, new HashSet<Integer>());
for(int[] edge : prerequisites) {
if(!adjList.get(edge[0]).contains(edge[1])) {
adjList.get(edge[0]).add(edge[1]);
indegree[edge[1]]++;
}
}
int topoVertical = findZeroIndegree(indegree);
while(topoVertical!=-1) {
indegree[topoVertical] = -1; // -1 stands for visited
for(Integer vertical : adjList.get(topoVertical)) {
indegree[vertical]--;
}
adjList.remove(topoVertical);
topoVertical = findZeroIndegree(indegree);
}
return isAllVisited(indegree);
}
private int findZeroIndegree(int[] indegree) {
for(int i = 0;i<indegree.length;i++)
if(indegree[i]==0)
return i;
return -1;
}
private boolean isAllVisited(int[] indegree) {
for(int i : indegree)
if(i!=-1)
return false;
return true;
}
}

成绩：

24ms,beats 47.84%，众数12ms,6.33%

Cmershen的碎碎念：

首先复习拓扑排序的定义和算法：对一个有向无环图(Directed Acyclic Graph简称DAG)G进行拓扑排序，是将G中所有顶点排成一个线性序列，使得图中任意一对顶点u和v，若边(u,v)∈E(G)，则u在线性序列中出现在v之前。通常，这样的线性序列称为满足拓扑次序(Topological Order)的序列，简称拓扑序列。

算法：

(1) 选择一个入度为0的顶点并输出之；

(2) 从网中删除此顶点及所有出边。

循环结束后，若输出的顶点数小于网中的顶点数，则输出“有回路”信息，否则输出的顶点序列就是一种拓扑序列。

而课程先修（或一个工程中各个子工程先后完成关系）是拓扑排序中最基本的问题。