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Diff for: ‎docs/0000-19-topological-sort.adoc

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[#0000-19-topological-sort]
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= Topological Sort (Graph) 拓扑排序
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拓扑排序模式用来寻找一种线性的顺序，这些元素之间具有依懒性。比如，如果事件B依赖于事件A，那A在拓扑排序顺序中排在B的前面。
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拓扑排序模式用来寻找一种线性的顺序，这些元素之间具有依懒性。比如，如果事件 B 依赖于事件 A，那 A 在拓扑排序顺序中排在 B 的前面。
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这种模式定义了一种简单方式来理解拓扑排序这种技术。
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这种模式是这样奏效的：
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. 初始化
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.. 借助于HashMap将图保存成邻接表形式。
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.. 找到所有的起点，用HashMap来帮助记录每个节点的入度
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.. 借助于 `Map` 将图保存成邻接表形式。
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.. 找到所有的起点，用 `Map` 来帮助记录每个节点的入度
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. 创建图，找到每个节点的入度
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.. 利用输入，把图建好，然后遍历一下图，将入度信息记录在HashMap中
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.. 利用输入，把图建好，然后遍历一下图，将入度信息记录在 `Map` 中
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. 找所有的起点
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.. 所有入度为0的节点，都是有效的起点，而且我们讲他们都加入到一个队列中
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.. 所有入度为 `0` 的节点，都是有效的起点，而且我们讲他们都加入到一个队列中
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. 排序
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.. 对每个起点，执行以下步骤
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... 把它加到结果的顺序中
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... 将其在图中的孩子节点取到
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... 将其孩子的入度减少1
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... 如果孩子的入度变为0，则改孩子节点成为起点，将其加入队列中
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.. 重复（a）过程，直到起点队列为空。
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.. 重复上述过程，直到起点队列为空。
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用一句话概括：将依赖关系转化成一张有向图，如果这张图中的节点没有循环依赖，那么则方案可行，否则方案不可行。
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TIP: 这里解释的是一种广度优先搜索，看 https://leetcode.cn/problems/course-schedule/solutions/359392/ke-cheng-biao-by-leetcode-solution/[207. 课程表 - 官方题解^]，还存在一种深度优先搜索的处理办法，有机会尝试一下。
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拓扑排序模式识别：
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. xref:0207-course-schedule.adoc[207. Course Schedule]
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. xref:0210-course-schedule-ii.adoc[210. Course Schedule II]
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. xref:0269-alien-dictionary.adoc[269. Alien Dictionary]
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. xref:0310-minimum-height-trees.adoc[310. Minimum Height Trees]
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. xref:0329-longest-increasing-path-in-a-matrix.adoc[329. Longest Increasing Path in a Matrix]
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. xref:0444-sequence-reconstruction.adoc[444. Sequence Reconstruction]
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. xref:0631-design-excel-sum-formula.adoc[631. Design Excel Sum Formula]
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. xref:0802-find-eventual-safe-states.adoc[802. Find Eventual Safe States]
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. xref:0851-loud-and-rich.adoc[851. Loud and Rich]
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. xref:0913-cat-and-mouse.adoc[913. Cat and Mouse]
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. xref:1059-all-paths-from-source-lead-to-destination.adoc[1059. All Paths from Source Lead to Destination]
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. xref:1136-parallel-courses.adoc[1136. Parallel Courses]
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. xref:1203-sort-items-by-groups-respecting-dependencies.adoc[1203. Sort Items by Groups Respecting Dependencies]
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. xref:1245-tree-diameter.adoc[1245. Tree Diameter]
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. xref:1462-course-schedule-iv.adoc[1462. Course Schedule IV]
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. xref:1591-strange-printer-ii.adoc[1591. Strange Printer II]
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. xref:1632-rank-transform-of-a-matrix.adoc[1632. Rank Transform of a Matrix]
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. xref:1728-cat-and-mouse-ii.adoc[1728. Cat and Mouse II]
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. xref:1786-number-of-restricted-paths-from-first-to-last-node.adoc[1786. Number of Restricted Paths From First to Last Node]
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. xref:1857-largest-color-value-in-a-directed-graph.adoc[1857. Largest Color Value in a Directed Graph]
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. xref:1916-count-ways-to-build-rooms-in-an-ant-colony.adoc[1916. Count Ways to Build Rooms in an Ant Colony]
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. xref:1976-number-of-ways-to-arrive-at-destination.adoc[1976. Number of Ways to Arrive at Destination]
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. xref:2050-parallel-courses-iii.adoc[2050. Parallel Courses III]
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. xref:2115-find-all-possible-recipes-from-given-supplies.adoc[2115. Find All Possible Recipes from Given Supplies]
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. xref:2127-maximum-employees-to-be-invited-to-a-meeting.adoc[2127. Maximum Employees to Be Invited to a Meeting]
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. xref:2192-all-ancestors-of-a-node-in-a-directed-acyclic-graph.adoc[2192. All Ancestors of a Node in a Directed Acyclic Graph]
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. xref:2246-longest-path-with-different-adjacent-characters.adoc[2246. Longest Path With Different Adjacent Characters]
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. xref:2328-number-of-increasing-paths-in-a-grid.adoc[2328. Number of Increasing Paths in a Grid]
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. xref:2360-longest-cycle-in-a-graph.adoc[2360. Longest Cycle in a Graph]
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. xref:2371-minimize-maximum-value-in-a-grid.adoc[2371. Minimize Maximum Value in a Grid]
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. xref:2392-build-a-matrix-with-conditions.adoc[2392. Build a Matrix With Conditions]
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. xref:2603-collect-coins-in-a-tree.adoc[2603. Collect Coins in a Tree]
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. xref:3383-minimum-runes-to-add-to-cast-spell.adoc[3383. Minimum Runes to Add to Cast Spell]
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. xref:3435-frequencies-of-shortest-supersequences.adoc[3435. Frequencies of Shortest Supersequences]
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. xref:3481-apply-substitutions.adoc[3481. Apply Substitutions]
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. xref:0310-minimum-height-trees.adoc[310. Minimum Height Trees]
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== 参考资料
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Diff for: ‎docs/0207-course-schedule.adoc

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[#0207-course-schedule]
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= 207. Course Schedule
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= 207. 课程表
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{leetcode}/problems/course-schedule/[LeetCode - Course Schedule^]
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https://leetcode.cn/problems/course-schedule/[LeetCode - 207. 课程表 ^]
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There are a total of _n_ courses you have to take, labeled from `0` to `n-1`.
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你这个学期必须选修 `numCourses` 门课程，记为 `0` 到 `numCourses - 1` 。
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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]`
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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] = [a~i~, b~i~]`，表示如果要学习课程 `a~i~` 则 *必须* 先学习课程 `b~i~`。
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Given the total number of courses and a list of prerequisite *pairs*, is it possible for you to finish all courses?
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* 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` 。
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*Example 1:*
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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` 。
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[subs="verbatim,quotes,macros"]
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----
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*Input:* 2, [[1,0]]
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*Output:* true
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*Explanation:* There are a total of 2 courses to take.
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To take course 1 you should have finished course 0. So it is possible.
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----
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*示例 1：*
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*Example 2:*
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....
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输入：numCourses = 2, prerequisites = [[1,0]]
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输出：true
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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。
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....
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[subs="verbatim,quotes,macros"]
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----
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*Input:* 2, [[1,0],[0,1]]
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*Output:* false
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*Explanation:* There are a total of 2 courses to take.
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To take course 1 you should have finished course 0, and to take course 0 you should
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also have finished course 1. So it is impossible.
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*示例 2：*
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----
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....
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输入：numCourses = 2, prerequisites = [[1,0],[0,1]]
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输出：false
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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。
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....
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*Note:*
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*提示：*
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. The input prerequisites is a graph represented by *a list of edges*, not adjacency matrices. Read more about https://www.khanacademy.org/computing/computer-science/algorithms/graph-representation/a/representing-graphs[how a graph is represented^].
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. You may assume that there are no duplicate edges in the input prerequisites.
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* `+1 <= numCourses <= 2000+`
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* `+0 <= prerequisites.length <= 5000+`
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* `prerequisites[i].length == 2`
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* `0 \<= a~i~, b~i~ < numCourses`
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* `prerequisites[i]` 中的所有课程对 *互不相同*
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== 思路分析
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TODO: 研究一下图相关算法和拓扑排序。
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拓扑排序通常用来“排序”具有依赖关系的任务。
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include::{sourcedir}/_0207_CourseSchedule_2.java[tag=answer]
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----
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--
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三刷::
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+
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--
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[{java_src_attr}]
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----
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include::{sourcedir}/_0207_CourseSchedule_3.java[tag=answer]
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----
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--
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====
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. https://www.cnblogs.com/bigsai/p/11489260.html[拓扑排序详解与实现^]
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. https://jingsam.github.io/2020/08/11/topological-sort.html[拓扑排序原理^]
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. https://leetcode.cn/problems/course-schedule/solutions/18806/course-schedule-tuo-bu-pai-xu-bfsdfsliang-chong-fa/?envType=study-plan-v2&envId=selected-coding-interview[207. 课程表 - 拓扑排序：入度表BFS法 / DFS法，清晰图解^]
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. https://leetcode.cn/problems/course-schedule/solutions/359392/ke-cheng-biao-by-leetcode-solution/?envType=study-plan-v2&envId=selected-coding-interview[207. 课程表 - 官方题解^]
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. https://leetcode.cn/problems/course-schedule/solutions/359392/ke-cheng-biao-by-leetcode-solution/[207. 课程表 - 官方题解^]
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. https://www.khanacademy.org/computing/computer-science/algorithms/graph-representation/a/representing-graphs[Representing graphs (article) | Algorithms | Khan Academy^]
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Diff for: ‎logbook/202503.adoc

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|{doc_base_url}/0015-3sum.adoc[题解]
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|✅ 利用递归“降维”
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|{counter:codes2503}
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|{leetcode_base_url}/course-schedule/[207. Course Schedule^]
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|{doc_base_url}/0207-course-schedule.adoc[题解]
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|✅ 拓扑排序
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|===
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截止目前，本轮练习一共完成 {codes2503} 道题。

Diff for: ‎src/main/java/com/diguage/algo/leetcode/_0207_CourseSchedule_3.java

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package com.diguage.algo.leetcode;
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import java.util.*;
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public class _0207_CourseSchedule_3 {
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// tag::answer[]
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/**
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* @author D瓜哥 · https://www.diguage.com
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* @since 2025-04-04 09:09:52
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*/
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public boolean canFinish(int numCourses, int[][] prerequisites) {
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Map<Integer, List<Integer>> graph = new HashMap<>();
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int[] indegree = new int[numCourses];
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for (int[] p : prerequisites) {
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List<Integer> children = graph.computeIfAbsent(p[0], k -> new ArrayList<>());
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children.add(p[1]);
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indegree[p[1]]++;
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}
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Queue<Integer> queue = new LinkedList<>();
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for (int i = 0; i < indegree.length; i++) {
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if (indegree[i] == 0) {
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queue.offer(i);
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}
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}
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int cnt = 0;
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while (!queue.isEmpty()) {
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Integer c = queue.poll();
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cnt++;
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List<Integer> children = graph.getOrDefault(c, Collections.emptyList());
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for (Integer child : children) {
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indegree[child]--;
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if (indegree[child] == 0) {
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queue.offer(child);
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}
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}
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}
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return cnt == numCourses;
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}
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// end::answer[]
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}

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