207. 课程表

English Version

题目描述

你这个学期必须选修 numCourses 门课程，记为 0 到 numCourses - 1 。

在选修某些课程之前需要一些先修课程。 先修课程按数组 prerequisites 给出，其中 prerequisites[i] = [ai, bi] ，表示如果要学习课程 ai 则 必须 先学习课程  bi 。

• 例如，先修课程对 [0, 1] 表示：想要学习课程 0 ，你需要先完成课程 1 。

请你判断是否可能完成所有课程的学习？如果可以，返回 true ；否则，返回 false 。

 

示例 1：

输入：numCourses = 2, prerequisites = [[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 。这是可能的。

示例 2：

输入：numCourses = 2, prerequisites = [[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：

• 1 <= numCourses <= 105
• 0 <= prerequisites.length <= 5000
• prerequisites[i].length == 2
• 0 <= ai, bi < numCourses
• prerequisites[i] 中的所有课程对 互不相同

解法

方法一：拓扑排序

对于本题，我们可以将课程看作图中的节点，先修课程看作图中的边，那么我们可以将本题转化为判断有向图中是否存在环。

具体地，我们可以使用拓扑排序的思想，对于每个入度为 $0$ 的节点，我们将其出度的节点的入度减 $1$，直到所有节点都被遍历到。

如果所有节点都被遍历到，说明图中不存在环，那么我们就可以完成所有课程的学习；否则，我们就无法完成所有课程的学习。

时间复杂度 $O(n+m)$，空间复杂度 $O(n+m)$。其中 $n$ 和 $m$ 分别为课程数和先修课程数。

Python3

class Solution:
    def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:
        g = defaultdict(list)
        indeg = [0] * numCourses
        for a, b in prerequisites:
            g[b].append(a)
            indeg[a] += 1
        cnt = 0
        q = deque(i for i, x in enumerate(indeg) if x == 0)
        while q:
            i = q.popleft()
            cnt += 1
            for j in g[i]:
                indeg[j] -= 1
                if indeg[j] == 0:
                    q.append(j)
        return cnt == numCourses

Java

class Solution {
    public boolean canFinish(int numCourses, int[][] prerequisites) {
        List<Integer>[] g = new List[numCourses];
        Arrays.setAll(g, k -> new ArrayList<>());
        int[] indeg = new int[numCourses];
        for (var p : prerequisites) {
            int a = p[0], b = p[1];
            g[b].add(a);
            ++indeg[a];
        }
        Deque<Integer> q = new ArrayDeque<>();
        for (int i = 0; i < numCourses; ++i) {
            if (indeg[i] == 0) {
                q.offer(i);
            }
        }
        int cnt = 0;
        while (!q.isEmpty()) {
            int i = q.poll();
            ++cnt;
            for (int j : g[i]) {
                if (--indeg[j] == 0) {
                    q.offer(j);
                }
            }
        }
        return cnt == numCourses;
    }
}

C++

class Solution {
public:
    bool canFinish(int numCourses, vector<vector<int>>& prerequisites) {
        vector<vector<int>> g(numCourses);
        vector<int> indeg(numCourses);
        for (auto& p : prerequisites) {
            int a = p[0], b = p[1];
            g[b].push_back(a);
            ++indeg[a];
        }
        queue<int> q;
        for (int i = 0; i < numCourses; ++i) {
            if (indeg[i] == 0) {
                q.push(i);
            }
        }
        int cnt = 0;
        while (!q.empty()) {
            int i = q.front();
            q.pop();
            ++cnt;
            for (int j : g[i]) {
                if (--indeg[j] == 0) {
                    q.push(j);
                }
            }
        }
        return cnt == numCourses;
    }
};

Rust

use std::collections::VecDeque;

impl Solution {
    #[allow(dead_code)]
    pub fn can_finish(num_course: i32, prerequisites: Vec<Vec<i32>>) -> bool {
        let num_course = num_course as usize;
        // The graph representation
        let mut graph: Vec<Vec<i32>> = vec![vec![]; num_course];
        // Record the in degree for each node
        let mut in_degree_vec: Vec<i32> = vec![0; num_course];
        let mut q: VecDeque<usize> = VecDeque::new();
        let mut count = 0;

        // Initialize the graph & in degree vector
        for p in &prerequisites {
            let (from, to) = (p[0], p[1]);
            graph[from as usize].push(to);
            in_degree_vec[to as usize] += 1;
        }

        // Enqueue the first batch of nodes with in degree 0
        for i in 0..num_course {
            if in_degree_vec[i] == 0 {
                q.push_back(i);
            }
        }

        // Begin the traverse & update through the graph
        while !q.is_empty() {
            // Get the current node index
            let index = q.front().unwrap().clone();
            // This course can be finished
            count += 1;
            q.pop_front();
            for i in &graph[index] {
                // Update the in degree for the current node
                in_degree_vec[*i as usize] -= 1;
                // See if can be enqueued
                if in_degree_vec[*i as usize] == 0 {
                    q.push_back(*i as usize);
                }
            }
        }

        count == num_course
    }
}

Go

func canFinish(numCourses int, prerequisites [][]int) bool {
	g := make([][]int, numCourses)
	indeg := make([]int, numCourses)
	for _, p := range prerequisites {
		a, b := p[0], p[1]
		g[b] = append(g[b], a)
		indeg[a]++
	}
	q := []int{}
	for i, x := range indeg {
		if x == 0 {
			q = append(q, i)
		}
	}
	cnt := 0
	for len(q) > 0 {
		i := q[0]
		q = q[1:]
		cnt++
		for _, j := range g[i] {
			indeg[j]--
			if indeg[j] == 0 {
				q = append(q, j)
			}
		}
	}
	return cnt == numCourses
}

TypeScript

function canFinish(numCourses: number, prerequisites: number[][]): boolean {
    const g: number[][] = new Array(numCourses).fill(0).map(() => []);
    const indeg: number[] = new Array(numCourses).fill(0);
    for (const [a, b] of prerequisites) {
        g[b].push(a);
        indeg[a]++;
    }
    const q: number[] = [];
    for (let i = 0; i < numCourses; ++i) {
        if (indeg[i] == 0) {
            q.push(i);
        }
    }
    let cnt = 0;
    while (q.length) {
        const i = q.shift()!;
        cnt++;
        for (const j of g[i]) {
            if (--indeg[j] == 0) {
                q.push(j);
            }
        }
    }
    return cnt == numCourses;
}

C#

public class Solution {
    public bool CanFinish(int numCourses, int[][] prerequisites) {
        var g = new List<int>[numCourses];
        for (int i = 0; i < numCourses; ++i) {
            g[i] = new List<int>();
        }
        var indeg = new int[numCourses];
        foreach (var p in prerequisites) {
            int a = p[0], b = p[1];
            g[b].Add(a);
            ++indeg[a];
        }
        var q = new Queue<int>();
        for (int i = 0; i < numCourses; ++i) {
            if (indeg[i] == 0) {
                q.Enqueue(i);
            }
        }
        var cnt = 0;
        while (q.Count > 0) {
            int i = q.Dequeue();
            ++cnt;
            foreach (int j in g[i]) {
                if (--indeg[j] == 0) {
                    q.Enqueue(j);
                }
            }
        }
        return cnt == numCourses;
    }
}

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