207 Course Schedule 課程表

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

題目描述:
你這個(gè)學(xué)期必須選修 numCourse 門(mén)課程，記為 0 到 numCourse-1 。

在選修某些課程之前需要一些先修課程。 例如，想要學(xué)習(xí)課程 0 ，你需要先完成課程 1 ，我們用一個(gè)匹配來(lái)表示他們：[0,1]

給定課程總量以及它們的先決條件，請(qǐng)你判斷是否可能完成所有課程的學(xué)習(xí)？

示例 :

示例 1:

輸入: 2, [[1,0]]
輸出: true
解釋: 總共有 2 門(mén)課程。學(xué)習(xí)課程 1 之前，你需要完成課程 0。所以這是可能的。

示例 2:

輸入: 2, [[1,0],[0,1]]
輸出: false
解釋: 總共有 2 門(mén)課程。學(xué)習(xí)課程 1 之前，你需要先完成?課程 0；并且學(xué)習(xí)課程 0 之前，你還應(yīng)先完成課程 1。這是不可能的。

提示：

輸入的先決條件是由 邊緣列表 表示的圖形，而不是 鄰接矩陣 。詳情請(qǐng)參見(jiàn)圖的表示法。
你可以假定輸入的先決條件中沒(méi)有重復(fù)的邊。
1 <= numCourses <= 10^5

思路:

拓?fù)渑判?br> 先將入度為 0的頂點(diǎn)加入一個(gè)處理隊(duì)列
每次取出入度為 0的頂點(diǎn), 將這個(gè)頂點(diǎn)的所有鄰接點(diǎn)入度 - 1, 再將入度為 0的點(diǎn)加入隊(duì)列
比較課程數(shù)與處理的頂點(diǎn)數(shù)是否相等
時(shí)間復(fù)雜度O(n + m), 空間復(fù)雜度O(n + m), 其中 n為課程數(shù), m為先修課程數(shù)

代碼:
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