207 Course Schedule 课程表
Description:
There are a total of numCourses courses you have to take, labeled from 0 to numCourses-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:
Example 1:
Input: numCourses = 2, prerequisites = [[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: numCourses = 2, prerequisites = [[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.
Constraints:
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.
1 <= numCourses <= 10^5
题目描述:
你这个学期必须选修 numCourse 门课程，记为 0 到 numCourse-1 。
在选修某些课程之前需要一些先修课程。 例如，想要学习课程 0 ，你需要先完成课程 1 ，我们用一个匹配来表示他们：[0,1]
给定课程总量以及它们的先决条件，请你判断是否可能完成所有课程的学习？
示例 :
示例 1:
输入: 2, [[1,0]]
输出: true
解释: 总共有 2 门课程。学习课程 1 之前，你需要完成课程 0。所以这是可能的。
示例 2:
输入: 2, [[1,0],[0,1]]
输出: false
解释: 总共有 2 门课程。学习课程 1 之前，你需要先完成课程 0；并且学习课程 0 之前，你还应先完成课程 1。这是不可能的。
提示：
输入的先决条件是由 边缘列表 表示的图形，而不是 邻接矩阵 。详情请参见图的表示法。
你可以假定输入的先决条件中没有重复的边。
1 <= numCourses <= 10^5
思路:
拓扑排序
先将入度为 0的顶点加入一个处理队列
每次取出入度为 0的顶点, 将这个顶点的所有邻接点入度 - 1, 再将入度为 0的点加入队列
比较课程数与处理的顶点数是否相等
时间复杂度O(n + m), 空间复杂度O(n + m), 其中 n为课程数, m为先修课程数
代码:
C++:
class Solution
{
public:
bool canFinish(int numCourses, vector<vector<int>>& prerequisites)
{
if (prerequisites.empty()) return true;
map<int, vector<int>> adjcent;
vector<int> indegree(numCourses, 0);
queue<int> q;
int count = 0;
for (auto& course : prerequisites)
{
adjcent[course[1]].push_back(course[0]);
++indegree[course[0]];
}
for (int i = 0; i < numCourses; i++) if (!indegree[i]) q.push(i);
while (q.size())
{
auto cur = q.front();
q.pop();
++count;
for (auto first : adjcent[cur])
{
--indegree[first];
if (!indegree[first]) q.push(first);
}
}
return count == numCourses;
}
};
Java:
class Solution {
public boolean canFinish(int numCourses, int[][] prerequisites) {
if (prerequisites == null || prerequisites.length == 0) return true;
int count = 0, indrgree[] = new int[numCourses];
List<List<Integer>> adjcent = new ArrayList<>(numCourses);
for (int i = 0; i < numCourses; i++) adjcent.add(new ArrayList<Integer>());
for (int i = 0; i < prerequisites.length; i++) {
++indrgree[prerequisites[i][1]];
adjcent.get(prerequisites[i][0]).add(prerequisites[i][1]);
}
Queue<Integer> queue = new LinkedList<>();
for (int i = 0; i < numCourses; i++) if (indrgree[i] == 0) queue.offer(i);
while (!queue.isEmpty()) {
int cur = queue.poll();
++count;
for (int first : adjcent.get(cur)) {
--indrgree[first];
if (indrgree[first] == 0) queue.offer(first);
}
}
return count == numCourses;
}
}
Python:
class Solution:
def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:
if not prerequisites:
return True
indegree, adjcent = [0] * numCourses, [set() for _ in range(numCourses)]
for second, first in prerequisites:
indegree[second] += 1
adjcent[first].add(second)
count, queue = 0, [i for i in range(numCourses) if not indegree[i]]
while queue:
cur = queue.pop(0)
count += 1
for first in adjcent[cur]:
indegree[first] -= 1
if not indegree[first]:
queue.append(first)
return count == numCourses
