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package class33;
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import java.util.ArrayList;
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import java.util.HashMap;
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import java.util.LinkedList;
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import java.util.Queue;
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public class Problem_0207_CourseSchedule {
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// 一个node，就是一个课程
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// name是课程的编号
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// in是课程的入度
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public static class Course {
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public int name;
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public int in;
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public ArrayList<Course> nexts;
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public Course(int n) {
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name = n;
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in = 0;
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nexts = new ArrayList<>();
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}
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}
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public static boolean canFinish1(int numCourses, int[][] prerequisites) {
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if (prerequisites == null || prerequisites.length == 0) {
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return true;
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}
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// 一个编号 对应 一个课的实例
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HashMap<Integer, Course> nodes = new HashMap<>();
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for (int[] arr : prerequisites) {
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int to = arr[0];
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int from = arr[1]; // from -> to
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if (!nodes.containsKey(to)) {
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nodes.put(to, new Course(to));
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}
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if (!nodes.containsKey(from)) {
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nodes.put(from, new Course(from));
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}
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Course t = nodes.get(to);
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Course f = nodes.get(from);
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f.nexts.add(t);
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t.in++;
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}
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int needPrerequisiteNums = nodes.size();
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Queue<Course> zeroInQueue = new LinkedList<>();
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for (Course node : nodes.values()) {
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if (node.in == 0) {
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zeroInQueue.add(node);
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}
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}
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int count = 0;
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while (!zeroInQueue.isEmpty()) {
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Course cur = zeroInQueue.poll();
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count++;
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for (Course next : cur.nexts) {
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if (--next.in == 0) {
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zeroInQueue.add(next);
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}
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}
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}
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return count == needPrerequisiteNums;
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}
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// 和方法1算法过程一样
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// 但是写法优化了
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public static boolean canFinish2(int courses, int[][] relation) {
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if (relation == null || relation.length == 0) {
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return true;
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}
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// 3 : 0 1 2
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// nexts : 0 {}
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// 1 {}
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// 2 {}
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// 3 {0,1,2}
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ArrayList<ArrayList<Integer>> nexts = new ArrayList<>();
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for (int i = 0; i < courses; i++) {
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nexts.add(new ArrayList<>());
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}
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// 3 入度 1 in[3] == 1
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int[] in = new int[courses];
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for (int[] arr : relation) {
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// arr[1] from arr[0] to
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nexts.get(arr[1]).add(arr[0]);
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in[arr[0]]++;
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}
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// 队列
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int[] zero = new int[courses];
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// 该队列有效范围是[l,r)
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// 新来的数，放哪？r位置，r++
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// 出队列的数，从哪拿？l位置，l++
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// l == r 队列无元素 l < r 队列有元素
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int l = 0;
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int r = 0;
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for (int i = 0; i < courses; i++) {
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if (in[i] == 0) {
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zero[r++] = i;
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}
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}
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int count = 0;
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while (l != r) {
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count++; // zero[l] 出队列 l++
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for (int next : nexts.get(zero[l++])) {
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if (--in[next] == 0) {
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zero[r++] = next;
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}
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}
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}
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return count == nexts.size();
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}
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}
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