leetcode207 Course Schedule

"""
There are a total of n courses you have to take, labeled from 0 to n-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 1:
Input: 2, [[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: 2, [[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.
"""
"""
此题是判断有向无环图
即拓扑排序
提供两种思路
第一种思路是维护一个indegrees数组
每次删除度数为0的结点，并将其指向结点度数减1
N次把所有结点取完 即返回True
"""
class Solution1(object):
    def canFinish(self, numCourses, prerequisites):
        from collections import defaultdict
        graph = defaultdict(list)    #defaultdict(<class 'list'>)
        indegrees = defaultdict(int) #defaultdict(<class 'int'>)
        for u, v in prerequisites:   #构建图{初始结点1:[目的结点1, 目的结点2, ...], ...}
            graph[v].append(u)       #结点度数{结点1:度数, 结点2:度数, ...}
            indegrees[u] += 1
        N = numCourses
        for i in range(N):           #每次循环将度为0的结点削掉，共N次
            zeroDegree = False
            for j in range(N):
                if indegrees[j] == 0: #找到度为0的结点
                    zeroDegree = True
                    break
            if not zeroDegree:
                return False
            indegrees[j] -= 1         #将度为0的结点减为-1
            for node in graph[j]:     #将其所连的结点各减1
                indegrees[node] -= 1
        return True

"""
第二种思路用DFS，递归，没有掌握
传送门：https://blog.csdn.net/fuxuemingzhu/article/details/82951771
"""
class Solution2(object):
    def canFinish(self, numCourses, prerequisites):
        from collections import defaultdict
        graph = defaultdict(list)  # defaultdict(<class 'list'>)
        for u, v in prerequisites: #构造图
            graph[v].append(u)
        N = numCourses
        visited = [0] * N
        # 0 = Unknown, 1 = visiting, 2 = visited
        for i in range(N):         #分别从N个结点出发调用dfs判断是否有环
            if not self.dfs(graph, visited, i):  #有环dfs返回True
                return False
        return True
    def dfs(self, graph, visited, i):   #这个函数云里雾里。。没懂
        if visited[i] == 1:
            return False
        if visited[i] == 2:
            return True
        visited[i] = 1
        for j in graph[i]:
            if not self.dfs(graph, visited, j):
                return False
        visited[i] = 2
        return True