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207. 课程表

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 。

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：

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上次编辑到这里，代码来自缓存 点击恢复默认模板
class Solution { public: bool canFinish(int numCourses, vector<vector<int>>& prerequisites) { } };

php 解法, 执行用时: 28 ms, 内存消耗: 22.5 MB, 提交时间: 2023-09-09 10:03:36 

class Solution {

    /**
     * @param Integer $numCourses
     * @param Integer[][] $prerequisites
     * @return Boolean
     */
    function canFinish($numCourses, $prerequisites) {
        $edges = array_fill(0, $numCourses, []);
        $indeg = array_fill(0, $numCourses, 0);

        foreach ( $prerequisites as $info ) {
            $edges[$info[1]][] = $info[0];
            $indeg[$info[0]]++;
        }
        
        $q = [];
        for ( $i = 0; $i < $numCourses; $i++ ) {
            if ( $indeg[$i] == 0 ) $q[] = $i;
        }
        $visited = 0;

        while ( !empty($q) ) {
            $visited++;
            $u = array_shift($q);
            foreach ( $edges[$u] as $v ) {
                $indeg[$v]--;
                if ( $indeg[$v] == 0 ) $q[] =  $v;
            }
        }

        return $visited == $numCourses;
    }
}

javascript 解法, 执行用时: 76 ms, 内存消耗: 45.2 MB, 提交时间: 2022-12-10 13:07:59 

/**
 * @param {number} numCourses
 * @param {number[][]} prerequisites
 * @return {boolean}
 */
const canFinish = (numCourses, prerequisites) => {
  const inDegree = new Array(numCourses).fill(0); // 入度数组
  const map = {};                                 // 邻接表
  for (let i = 0; i < prerequisites.length; i++) {
    inDegree[prerequisites[i][0]]++;              // 求课的初始入度值
    if (map[prerequisites[i][1]]) {               // 当前课已经存在于邻接表
      map[prerequisites[i][1]].push(prerequisites[i][0]); // 添加依赖它的后续课
    } else {                                      // 当前课不存在于邻接表
      map[prerequisites[i][1]] = [prerequisites[i][0]];
    }
  }
  const queue = [];
  for (let i = 0; i < inDegree.length; i++) { // 所有入度为0的课入列
    if (inDegree[i] == 0) queue.push(i);
  }
  let count = 0;
  while (queue.length) {
    const selected = queue.shift();           // 当前选的课，出列
    count++;                                  // 选课数+1
    const toEnQueue = map[selected];          // 获取这门课对应的后续课
    if (toEnQueue && toEnQueue.length) {      // 确实有后续课
      for (let i = 0; i < toEnQueue.length; i++) {
        inDegree[toEnQueue[i]]--;             // 依赖它的后续课的入度-1
        if (inDegree[toEnQueue[i]] == 0) {    // 如果因此减为0，入列
          queue.push(toEnQueue[i]);
        }
      }
    }
  }
  return count == numCourses; // 选了的课等于总课数，true，否则false
};

python3 解法, 执行用时: 48 ms, 内存消耗: 15.9 MB, 提交时间: 2022-12-10 13:07:35 

class Solution:
    def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:
        edges = collections.defaultdict(list)
        indeg = [0] * numCourses

        for info in prerequisites:
            edges[info[1]].append(info[0])
            indeg[info[0]] += 1
        
        q = collections.deque([u for u in range(numCourses) if indeg[u] == 0])
        visited = 0

        while q:
            visited += 1
            u = q.popleft()
            for v in edges[u]:
                indeg[v] -= 1
                if indeg[v] == 0:
                    q.append(v)

        return visited == numCourses

java 解法, 执行用时: 5 ms, 内存消耗: 41.4 MB, 提交时间: 2022-12-10 13:07:23 

class Solution {
    List<List<Integer>> edges;
    int[] indeg;

    public boolean canFinish(int numCourses, int[][] prerequisites) {
        edges = new ArrayList<List<Integer>>();
        for (int i = 0; i < numCourses; ++i) {
            edges.add(new ArrayList<Integer>());
        }
        indeg = new int[numCourses];
        for (int[] info : prerequisites) {
            edges.get(info[1]).add(info[0]);
            ++indeg[info[0]];
        }

        Queue<Integer> queue = new LinkedList<Integer>();
        for (int i = 0; i < numCourses; ++i) {
            if (indeg[i] == 0) {
                queue.offer(i);
            }
        }

        int visited = 0;
        while (!queue.isEmpty()) {
            ++visited;
            int u = queue.poll();
            for (int v: edges.get(u)) {
                --indeg[v];
                if (indeg[v] == 0) {
                    queue.offer(v);
                }
            }
        }

        return visited == numCourses;
    }
}

golang 解法, 执行用时: 8 ms, 内存消耗: 6 MB, 提交时间: 2022-12-10 13:07:10 

// bfs
func canFinish(numCourses int, prerequisites [][]int) bool {
    var (
        edges = make([][]int, numCourses)
        indeg = make([]int, numCourses)
        result []int
    )

    for _, info := range prerequisites {
        edges[info[1]] = append(edges[info[1]], info[0])
        indeg[info[0]]++
    }

    q := []int{}
    for i := 0; i < numCourses; i++ {
        if indeg[i] == 0 {
            q = append(q, i)
        }
    }

    for len(q) > 0 {
        u := q[0]
        q = q[1:]
        result = append(result, u)
        for _, v := range edges[u] {
            indeg[v]--
            if indeg[v] == 0 {
                q = append(q, v)
            }
        }
    }
    return len(result) == numCourses
}

golang 解法, 执行用时: 8 ms, 内存消耗: 6.6 MB, 提交时间: 2022-12-10 13:06:54 

func canFinish(numCourses int, prerequisites [][]int) bool {
    var (
        edges = make([][]int, numCourses)
        visited = make([]int, numCourses)
        result []int
        valid = true
        dfs func(u int)
    )

    dfs = func(u int) {
        visited[u] = 1
        for _, v := range edges[u] {
            if visited[v] == 0 {
                dfs(v)
                if !valid {
                    return
                }
            } else if visited[v] == 1 {
                valid = false
                return
            }
        }
        visited[u] = 2
        result = append(result, u)
    }

    for _, info := range prerequisites {
        edges[info[1]] = append(edges[info[1]], info[0])
    }

    for i := 0; i < numCourses && valid; i++ {
        if visited[i] == 0 {
            dfs(i)
        }
    }
    return valid
}

python3 解法, 执行用时: 44 ms, 内存消耗: 17.7 MB, 提交时间: 2022-12-10 13:06:41 

class Solution:
    def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:
        edges = collections.defaultdict(list)
        visited = [0] * numCourses
        result = list()
        valid = True

        for info in prerequisites:
            edges[info[1]].append(info[0])
        
        def dfs(u: int):
            nonlocal valid
            visited[u] = 1
            for v in edges[u]:
                if visited[v] == 0:
                    dfs(v)
                    if not valid:
                        return
                elif visited[v] == 1:
                    valid = False
                    return
            visited[u] = 2
            result.append(u)
        
        for i in range(numCourses):
            if valid and not visited[i]:
                dfs(i)
        
        return valid

java 解法, 执行用时: 3 ms, 内存消耗: 41.8 MB, 提交时间: 2022-12-10 13:06:27 

// dfs
class Solution {
    List<List<Integer>> edges;
    int[] visited;
    boolean valid = true;

    public boolean canFinish(int numCourses, int[][] prerequisites) {
        edges = new ArrayList<List<Integer>>();
        for (int i = 0; i < numCourses; ++i) {
            edges.add(new ArrayList<Integer>());
        }
        visited = new int[numCourses];
        for (int[] info : prerequisites) {
            edges.get(info[1]).add(info[0]);
        }
        for (int i = 0; i < numCourses && valid; ++i) {
            if (visited[i] == 0) {
                dfs(i);
            }
        }
        return valid;
    }

    public void dfs(int u) {
        visited[u] = 1;
        for (int v: edges.get(u)) {
            if (visited[v] == 0) {
                dfs(v);
                if (!valid) {
                    return;
                }
            } else if (visited[v] == 1) {
                valid = false;
                return;
            }
        }
        visited[u] = 2;
    }
}

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