Course Schedule Ii

0210.Course-Schedule-II/memo.md

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+# 210. Course Schedule II
+
+## step1
+最初集合で先祖を管理しようとしたが、閉路検出ができないことに気づき、statusを使う方針に書き換えた。
+19mほど。
+
+## 他の人のコードなど
+
+> DFSのトポロジカルソートですが、preorder（行きがけ）でTEMPORARYとマークか、すでにTEMPORARYなら、サイクルと判定。postorder（帰りがけ）でPERMANENTとマークするみたいな感じですね。
+
+https://github.com/potrue/leetcode/pull/77/changes
+
+入次数で管理するやり方。Kahnのアルゴリズムというものだった。他の問題で見た覚えがあるが使えなかった。
+
+https://ja.wikipedia.org/wiki/%E3%83%88%E3%83%9D%E3%83%AD%E3%82%B8%E3%82%AB%E3%83%AB%E3%82%BD%E3%83%BC%E3%83%88
+
+Kahnのアルゴリズムを書くのはBFS, DFSどちらでも可能だがBFSで実装する。
+
+## step2
+変数名を具体的にする

0210.Course-Schedule-II/step1_dfs.py

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+class Solution:
+    def findOrder(self, numCourses: int, prerequisites: list[list[int]]) -> list[int]:
+        adjacent_list = [[] for _ in range(numCourses)]
+        for second, first in prerequisites:
+            adjacent_list[first].append(second)
+
+        order = []
+        # 0: 未訪問, 1: 訪問中, 2: 訪問ずみ
+        status = [0] * numCourses
+
+        def dfs(i):
+            if status[i] == 1:
+                return False
+            if status[i] == 2:
+                return True
+
+            status[i] = 1
+            for j in adjacent_list[i]:
+                can_be_finished = dfs(j)
+                if not can_be_finished:
+                    return False
+
+            status[i] = 2
+            order.append(i)
+            return True
+
+        for i in range(numCourses):
+            if status[i] != 0:
+                continue
+            can_be_finished = dfs(i)
+            if not can_be_finished:
+                return []
+
+        return order[::-1]

0210.Course-Schedule-II/step2_dfs_revised.py

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+class Solution:
+    def findOrder(self, numCourses: int, prerequisites: list[list[int]]) -> list[int]:
+        next_courses = [[] for _ in range(numCourses)]
+        for second, first in prerequisites:
+            next_courses[first].append(second)
+
+        order = []
+        # 0: 未訪問, 1: 訪問中, 2: 訪問ずみ
+        status = [0] * numCourses
+
+        def dfs(course):
+            if status[course] == 1:
+                return False
+            if status[course] == 2:
+                return True
+
+            status[course] = 1
+            for next_course in next_courses[course]:
+                can_be_finished = dfs(next_course)
+                if not can_be_finished:
+                    return False
+
+            status[course] = 2
+            order.append(course)
+            return True
+
+        for course in range(numCourses):
+            if status[course] != 0:
+                continue
+            can_be_finished = dfs(course)
+            if not can_be_finished:
+                return []
+
+        return order[::-1]

0210.Course-Schedule-II/step2_kahn.py

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+import collections
+
+class Solution:
+    def findOrder(self, numCourses: int, prerequisites: list[list[int]]) -> list[int]:
+        adjacent_list = [[] for _ in range(numCourses)]
+        in_degree = [0] * numCourses
+        for second, first in prerequisites:
+            adjacent_list[first].append(second)
+            in_degree[second] += 1
+
+        queue = collections.deque()
+        for i in range(numCourses):
+            if in_degree[i] == 0:
+                queue.append(i)
+
+        order = []
+        while queue:
+            i = queue.popleft()
+            order.append(i)
+            for j in adjacent_list[i]:
+                in_degree[j] -= 1
+                if in_degree[j] == 0:
+                    queue.append(j)
+
+        if len(order) != numCourses:
+            return []
+        return order
+
+

0210.Course-Schedule-II/step2_kahn_revised.py

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+import collections
+
+class Solution:
+    def findOrder(self, numCourses: int, prerequisites: list[list[int]]) -> list[int]:
+        next_courses = [[] for _ in range(numCourses)]
+        in_degree = [0] * numCourses
+        for second, first in prerequisites:
+            next_courses[first].append(second)
+            in_degree[second] += 1
+
+        available_courses = collections.deque()
+        for course in range(numCourses):
+            if in_degree[course] == 0:
+                available_courses.append(course)
+
+        order = []
+        while available_courses:
+            course = available_courses.popleft()
+            order.append(course)
+            for next_course in next_courses[course]:
+                in_degree[next_course] -= 1
+                if in_degree[next_course] == 0:
+                    available_courses.append(next_course)
+
+        if len(order) != numCourses:
+            return []
+        return order
+
+