leetcode-210-课程表②

题目描述：

 

 

 

 第一次提交：

class Solution:
    def findOrder(self, numCourses: int, prerequisites: List[List[int]]) -> List[int]:
        indegree = [0 for _ in range(numCourses)]
        adj = [[] for _ in range(numCourses)]
        for cur,pre in prerequisites:
            adj[pre].append(cur)
            indegree[cur] += 1
        res = []
        queue = []
        for i in range(numCourses):
            if indegree[i] == 0:
                queue.append(i)
        while queue:
            i = queue.pop(0)
            res.append(i)
            for j in adj[i]:
                indegree[j] -= 1
                if indegree[j] == 0:
                    queue.append(j)
        return res if len(res) == numCourses else []

方法二：dfs + 栈 

class Solution:
    def findOrder(self, numCourses: int, prerequisites: List[List[int]]) -> List[int]:
        def dfs(i, adjacency, flags):
            if flags[i] == -1: return True
            if flags[i] == 1: return False
            flags[i] = 1
            for j in adjacency[i]:
                if not dfs(j, adjacency, flags): return False
            flags[i] = -1
            res.append(i)
            return True
        res = []
        adjacency = [[] for _ in range(numCourses)]
        flags = [0 for _ in range(numCourses)]
        for cur, pre in prerequisites:
            adjacency[pre].append(cur)
        for i in range(numCourses):
            if not dfs(i, adjacency, flags): return []
        return res[::-1]