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LeetCode-207-Course-Schedule.java
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LeetCode-207-Course-Schedule.java
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/*
This problem can be convert to check if there is cycle in graph.
1.DFS
http://blog.welkinlan.com/2015/05/09/course-schedule-leetcode-java-dfs/
Step1. Build graph. Put graph into HashMap<Integer, List<Integer>>
Step2. Run DFS to detect cycle
- Using a int[] to mark course status: 0 not visited, 1 visited(used in DFS), -1 finish visited
- Recursive to detect cycle. During recursive, if we find some course has already been visited, return true(contains cycle)
2.DFS(Using Matrix to store Graph)
TLE(Too many unuseful information, I think.)
2.BFS
https://leetcode.com/discuss/39456/java-dfs-and-bfs-solution
*/
public class Solution {
// 1.DFS(HashMap to store graph)
// public boolean canFinish(int numCourses, int[][] prerequisites) {
// if(prerequisites == null || prerequisites.length == 0) return true;
// // Construct graph
// HashMap<Integer, List<Integer>> map = new HashMap<Integer, List<Integer>>();
// for(int i = 0; i < prerequisites.length; i++){
// int from = prerequisites[i][1];
// int to = prerequisites[i][0];
// if(!map.containsKey(from)){
// map.put(from, new ArrayList<Integer>());
// }
// List<Integer> list = map.get(from);
// list.add(to);
// map.put(from, list);
// }
// // detect cycle one by one using DFS: not visited && has cycle
// int[] visited = new int[numCourses];
// for(int from = 0; from < numCourses; from++){
// if(visited[from] == 0 && hasCycle(from, visited, map)){
// return false;
// }
// }
// return true;
// }
// private boolean hasCycle(int from, int[] visited, HashMap<Integer, List<Integer>> map){
// if(visited[from] == 1) return true; //from has already visited
// visited[from] = 1;
// // means from is several other courses' prerequisite
// if(map.containsKey(from)){
// for(Integer to : map.get(from)){
// if(visited[to] == 1) return true; //to has already visited
// if(hasCycle(to, visited, map)) return true; //to is in cycle
// }
// }
// visited[from] = -1; // finish visit
// return false;
// }
// 2.DFS(Using Matrix to store Graph)
// public boolean canFinish(int numCourses, int[][] prerequisites) {
// if(prerequisites == null || prerequisites.length == 0) return true;
// // construct graph
// int[][] graph = new int[numCourses][numCourses];
// for(int i = 0; i < prerequisites.length; i++){
// int from = prerequisites[i][1];
// int to = prerequisites[i][0];
// graph[from][to] = 1;
// }
// // detect cycle recursive
// int[] visited = new int[numCourses];
// for(int from = 0; from < numCourses; from++){
// if(visited[from] == 0 && hasCycle(from, graph, visited)){
// return false; //there is cycle
// }
// }
// return true;
// }
// private boolean hasCycle(int from, int[][] graph, int[] visited){
// if(visited[from] == 1) return true; //has already visited from, there is cycle
// visited[from] = 1;
// for(int to = 0; to < graph[0].length; to++){
// if(graph[from][to] == 1){
// if(visited[to] == 1) return true; //has already visited
// if(hasCycle(to, graph, visited)) return true;
// }
// }
// visited[from] = -1; //finish visited
// return false;
// }
// 3.BFS
public boolean canFinish(int numCourses, int[][] prerequisites) {
if(prerequisites == null || prerequisites.length == 0) return true;
HashMap<Integer, List<Integer>> graph = new HashMap<Integer, List<Integer>>();
int[] indegree = new int[numCourses];
for(int i = 0; i < prerequisites.length; i++){
int from = prerequisites[i][1];
int to = prerequisites[i][0];
if(!graph.containsKey(from)){
graph.put(from, new LinkedList<Integer>());
}
indegree[to]++; // means we have one node could reach to, indegree
List<Integer> list = graph.get(from);
list.add(to);
graph.put(from, list);
}
// put those indegree = 0, into queue
Queue<Integer> queue = new LinkedList<Integer>();
for(int i = 0; i < numCourses; i++){
if(indegree[i] == 0){
queue.offer(i);
}
}
int count = 0;
while(!queue.isEmpty()){
int curr = queue.poll();
if(graph.containsKey(curr)){
for(int i : graph.get(curr)){
indegree[i]--; // minus 1 indegree
if(indegree[i] == 0){
queue.offer(i);
}
}
}
count++;
}
return count == numCourses;
}
}