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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?
Example 1:
Input: 2, [[1,0]] Output: true Explanation: There are a total of 2 courses to take. To take course 1 you should have finished course 0. So it is possible.
Example 2:
Input: 2, [[1,0],[0,1]] Output: false Explanation: 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.
- You may assume that there are no duplicate edges in the input prerequisites.
题目链接：https://leetcode.com/problems/course-schedule/
题目分析：同210
class Solution {
public boolean canFinish(int numCourses, int[][] prerequisites) {
int[] queue = new int[numCourses];
int[] indegree = new int[numCourses];
int head = 0, cnt = 0, enqueueCnt = 0;
Map<Integer, List<Integer>> g = new HashMap<>();
for (int i = 0; i < prerequisites.length; i++) {
indegree[prerequisites[i][0]]++;
List<Integer> list = g.get(prerequisites[i][1]);
if (list == null) {
list = new ArrayList<>();
list.add(prerequisites[i][0]);
g.put(prerequisites[i][1], list);
} else {
list.add(prerequisites[i][0]);
}
}
for (int i = 0; i < numCourses; i++) {
if (indegree[i] == 0) {
queue[head++] = i;
enqueueCnt++;
}
}
while (head > 0) {
int u = queue[--head];
List<Integer> linkNodes = g.get(u);
if (linkNodes != null) {
for (int i = 0; i < linkNodes.size(); i++) {
int v = linkNodes.get(i);
indegree[v]--;
if (indegree[v] == 0) {
queue[head++] = v;
enqueueCnt++;
}
}
}
}
return enqueueCnt == numCourses;
}
}