Course Schedule I II LeetCode Java

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.

这道题目考的是拓扑排序相关内容，不了解什么是拓扑排序，可以参考博客：/article/2388810.html

所以，这道题目也有两种解法(程序参考自：https://leetcode.com/discuss/39456/java-dfs-and-bfs-solution)：

BFS：

public class Solution {
public boolean canFinish(int numCourses, int[][] prerequisites) {
ArrayList[] graph = new ArrayList[numCourses];
int[] degree = new int[numCourses];
Queue queue = new LinkedList();
int count=0;

for(int i=0;i<numCourses;i++)
graph[i] = new ArrayList();

for(int i=0; i<prerequisites.length;i++){
degree[prerequisites[i][1]]++;
graph[prerequisites[i][0]].add(prerequisites[i][1]);
}
for(int i=0; i<degree.length;i++){
if(degree[i] == 0){
queue.add(i);
count++;
}
}

while(queue.size() != 0){
int course = (int)queue.poll();
for(int i=0; i<graph[course].size();i++){
int pointer = (int)graph[course].get(i);
degree[pointer]--;
if(degree[pointer] == 0){
queue.add(pointer);
count++;
}
}
}
if(count == numCourses)
return true;
else
return false;
}
}

DFS:

public boolean canFinishDFS(int numCourses, int[][] prerequisites) {
ArrayList[] graph = new ArrayList[numCourses];
for(int i=0;i<numCourses;i++)
graph[i] = new ArrayList();

boolean[] visited = new boolean[numCourses];
boolean[] visiting=new boolean[numCourses];
for(int i=0; i<prerequisites.length;i++){
graph[prerequisites[i][1]].add(prerequisites[i][0]);
}

for(int i=0; i<numCourses; i++){
if(!dfs(graph,visited,visiting,i))
return false;
}
return true;
}

private boolean dfs(ArrayList[] graph, boolean[] visited, boolean[] visiting, int course){
if(visiting[course])
return false;
if(visited[course])
return true;

visiting[course] = true;;

for(int i=0; i<graph[course].size();i++){
if(!dfs(graph,visited,visiting,(int)graph[course].get(i)))
return false;
}
visiting[course]=false;
visited[course] = true;
return true;
}

Course Schedule II

这道题目和上道题目思路类似，只不过要求输出选课的顺序

同样有两种解法

BFS：

public int[] findOrder(int numCourses, int[][] prerequisites){
ArrayList<Integer>[] graph=new ArrayList[numCourses];
int degree[] =new int[numCourses];
Queue<Integer> queue=new LinkedList<Integer>();
int count=0;
int[] rst=new int[numCourses];

for(int i=0;i<numCourses;i++){
graph[i]=new ArrayList<Integer>();
}

for(int i=0;i<prerequisites.length;i++){
degree[prerequisites[i][0]]++;
graph[prerequisites[i][1]].add(prerequisites[i][0]);
}

for(int i=0;i<degree.length;i++){
if(degree[i]==0){
queue.add(i);
rst[count]=i;
count++;
}
}
while(queue.size()!=0){
int course=(int)queue.poll();
for(int i=0;i<graph[course].size();i++){
int pointer=(int)graph[course].get(i);
degree[pointer]--;
if(degree[pointer]==0){
queue.add(pointer);
rst[count]=pointer;
count++;
}
}
}
if(count == numCourses)
return rst;
else
return new int[0];
}

DFS：

public class Solution {
private int[] rst;
private int count;
public int[] findOrder(int numCourses, int[][] prerequisites) {
ArrayList[] graph = new ArrayList[numCourses];
rst=new int[numCourses];
count=0;
for(int i=0;i<numCourses;i++)
graph[i] = new ArrayList();

boolean[] visited = new boolean[numCourses];
boolean[] visiting=new boolean[numCourses];
for(int i=0; i<prerequisites.length;i++){
graph[prerequisites[i][0]].add(prerequisites[i][1]);
}

for(int i=0; i<numCourses; i++){
if(!dfs2(graph,visited,visiting,i))
return new int[0];
}
return rst;
}

private boolean dfs2(ArrayList[] graph, boolean[] visited, boolean[] visiting, int course){
if(visiting[course])
return false;
if(visited[course])
return true;

visiting[course] = true;;

for(int i=0; i<graph[course].size();i++){
if(!dfs2(graph,visited,visiting,(int)graph[course].get(i)))
return false;
}
visiting[course]=false;
visited[course] = true;
rst[count]=course;
count++;
return true;
}
}