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.
首先根据题意构造一个图，然后用DFS判断图中是否存在环，如果存在环就不能完成课程。代码如下：
public class Solution {
boolean[] onStack;
boolean[] isVisited;
public boolean canFinish(int numCourses, int[][] prerequisites) {
if(prerequisites == null || prerequisites.length == 0) return true;
List<Integer>[] graph = new List[numCourses];
onStack = new boolean[numCourses];
isVisited = new boolean[numCourses];
for(int i = 0; i < prerequisites.length; i++) {
int key = prerequisites[i][1];
int value = prerequisites[i][0];
if(graph[key] == null) graph[key] = new ArrayList<Integer>();
graph[key].add(value);
}
for(int i = 0; i < numCourses; i++) {
if(isVisited[i] == false && hasCycle(i, graph) == true)
return false;
}
return true;
}
public boolean hasCycle(int i, List<Integer>[] graph) {
boolean cycle = false;
onStack[i] = true;
isVisited[i] = true;
if(graph[i] != null) {
for(int j : graph[i]) {
if(isVisited[j] == false) {
cycle = hasCycle(j, graph);
if(cycle == true)
break;
} else if(onStack[j] == true) {
cycle = true;
break;
}
}
}
onStack[i] = false;
return cycle;
}
}