210. Course Schedule II
There are a total of numCourses courses you have to take, labeled from 0 to numCourses - 1. You are given an array prerequisites where prerequisites[i] = [ai, bi] indicates that you must take course bi first if you want to take course ai.
- For example, the pair
[0, 1], indicates that to take course0you have to first take course1.
Return the ordering of courses you should take to finish all courses. If there are many valid answers, return any of them. If it is impossible to finish all courses, return an empty array.
Example 1:
Input: numCourses = 2, prerequisites = [[1,0]]
Output: [0,1]
Explanation: There are a total of 2 courses to take. To take course 1 you should have finished course 0. So the correct course order is [0,1].
Example 2:
Input: numCourses = 4, prerequisites = [[1,0],[2,0],[3,1],[3,2]]
Output: [0,2,1,3]
Explanation: There are a total of 4 courses to take. To take course 3 you should have finished both courses 1 and 2. Both courses 1 and 2 should be taken after you finished course 0.
So one correct course order is [0,1,2,3]. Another correct ordering is [0,2,1,3].
Example 3:
Solution:
这个问题是拓扑排序问题可以用dfs 和
class Solution {
public int[] findOrder(int numCourses, int[][] prerequisites) {
int[] result = new int[numCourses];
Map<Integer, List<Integer>> graph = new HashMap<>();
for (int i = 0; i < numCourses; i++){
graph.put(i, new ArrayList<>());
}
for (int[] prerequisite : prerequisites){
int course = prerequisite[0];
int pre = prerequisite[1];
graph.get(course).add(pre);
}
Set<Integer> visited = new HashSet<>();
Set<Integer> curPath = new HashSet<>();
List<Integer> subResult = new ArrayList<>();
for (int i = 0; i < numCourses; i++){
if (dfs(i, graph, visited, curPath, subResult) == true){
return new int[]{};
}
}
for (int i = 0; i < subResult.size(); i++){
result[i] = subResult.get(i);
}
return result;
}
private boolean dfs(int course, Map<Integer, List<Integer>> graph, Set<Integer> visited, Set<Integer> curPath, List<Integer> subResult){
if (curPath.contains(course)){
return true;
}
if (visited.contains(course)){
return false;
}
curPath.add(course);
for (int next : graph.get(course)){
if (dfs(next, graph, visited, curPath, subResult) == true){
return true;
}
}
curPath.remove(course);
visited.add(course);
subResult.add(course);
return false;
}
}
// TC: O(V+E)
// SC: O(V + E)
class Solution {
public int[] findOrder(int numCourses, int[][] prerequisites) {
int[] ans = new int[numCourses]; // [0, 1, 2, 3]
Map<Integer, List<Integer>> preMap = new HashMap<>();
// 0, []
// 1, [0]
// 2, [0]
// 3, [1,2]
for (int i = 0; i < numCourses; i++){
preMap.put(i, new ArrayList<Integer>());
}
for (int[] row : prerequisites){// [1,0]//[2,0] //[3,1]// [3,2]
int course = row[0]; // 1 // 2 // 3
int prereq = row[1];// 0 // 0 // 1
// List<Integer> allPrereq = preMap.get(course); // 1,[]
// allPrereq.add(prereq);// 1, [0]
// preMap.put(course, allPrereq); // 1,[0]
preMap.get(course).add(prereq); //
}
// 0, []
// 1, [0]
// 2, [0]
// 3, [1,2]
Set<Integer> visited = new HashSet<>();
// Set to keep track of visited nodes
Set<Integer> cycle = new HashSet<>();
// Set to detect cycles in the graph.
List<Integer> ansList = new ArrayList<>();
// List to store the courses in the order of completion.
for (int i = 0; i < numCourses; i++){
if (!dfs(i, preMap, ansList, visited, cycle)){
return new int[]{};
}
}
for (int i = 0; i < ansList.size(); i++){
ans[i] = ansList.get(i);
}
return ans;
}
public boolean dfs(int course, Map<Integer, List<Integer>> preMap, List<Integer> ansList, Set<Integer> visited, Set<Integer> cycle){
if (cycle.contains(course)){
return false;
}
// If the course is in the cycle set, we found a cycle, return false.
if (visited.contains(course)){
return true;
}
// If the course is already visited, no need to process it again.
cycle.add(course);
for(int prereq : preMap.get(course)){
if (!dfs(prereq, preMap, ansList, visited, cycle)){
return false;
}
}
// Recursively perform DFS on all prerequisites of the course.
cycle.remove(course);
// Remove the course from the cycle set as we're done processing it.
visited.add(course);
// Mark the course as visited.
ansList.add(course);
// Return true as the course can be completed.
return true;
}
}
// TC: O(V+E)
// SC: O(V+E)
Kahn's Algorithm
基于(BFS)
核心思想: 做完一件事打勾
步骤:
- 统计所有点的入度
- 执行入度为0的点代表的任务
- 每执行完一个任务, 所有后继任务的入度减1
- 重复2和3直到所有可执行任务都被执行完
- 如果还有剩余任务, 返回无解, 否则返回任务执行顺序
class Solution {
public int[] findOrder(int numCourses, int[][] prerequisites) {
// 创建图和入度数组
Map<Integer, List<Integer>> graph = new HashMap<>();
int[] inDegree = new int[numCourses];
int[] result = new int[numCourses]; // 用于存储拓扑排序结果
// 初始化图结构
for (int i = 0; i < numCourses; i++) {
graph.put(i, new ArrayList<>());
}
// 填充图和计算入度
for (int[] prerequisite : prerequisites) {
int course = prerequisite[0];
int pre = prerequisite[1];
graph.get(pre).add(course); // pre 依赖 course
inDegree[course]++; // 记录 course 的入度
}
// 使用队列存储所有入度为 0 的节点
Queue<Integer> queue = new LinkedList<>();
for (int i = 0; i < numCourses; i++) {
if (inDegree[i] == 0) {
queue.offer(i); // 入度为 0 的节点可以作为起点
}
}
int index = 0; // 记录排序位置
// 进行拓扑排序
while (!queue.isEmpty()) {
int course = queue.poll();
result[index++] = course; // 将课程添加到结果数组中
// 遍历该课程的邻居节点
for (int neighbor : graph.get(course)) {
inDegree[neighbor]--;
if (inDegree[neighbor] == 0) {
queue.offer(neighbor); // 入度为 0 的邻居节点可以入队
}
}
}
// 如果结果中的课程数量不等于 numCourses，说明存在环，无法完成所有课程
if (index != numCourses) {
return new int[0]; // 返回空数组表示不能完成所有课程
}
return result; // 返回拓扑排序结果
}
}
https://www.youtube.com/watch?v=Akt3glAwyfY&t=145s
https://www.bilibili.com/video/BV1PZ4y1X7w4/?spm_id_from=333.337.search-card.all.click&vd_source=73e7d2c4251a7c9000b22d21b70f5635