0210. Course Schedule II (M)
Course Schedule II (M)
题目
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, return the ordering of courses you should take to finish all courses.
There may be multiple correct orders, you just need to return one of them. If it is impossible to finish all courses, return an empty array.
Example 1:
Input: 2, [[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: 4, [[1,0],[2,0],[3,1],[3,2]]
Output: [0,1,2,3] or [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] .
Note:
- The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.
- You may assume that there are no duplicate edges in the input prerequisites.
题意
给出一组前置课程信息，在选课程x前，必须已经修完x所有的前置课程。要求给出一组能够修完所有课程的选课顺序，如果找不出则返回空数组。
思路
典型的拓扑排序问题，要求找到给定图的拓扑序列，如果存在环则返回空数组。求拓扑序列有两种方法：postOrder以及入度法。
代码实现
Java
postOrder
class Solution {
public int[] findOrder(int numCourses, int[][] prerequisites) {
List<Integer> temp = new ArrayList<>();
List<List<Integer>> graph = new ArrayList<>();
// 生成图，注意图中的边是由前置课程指向当前课程
for (int i = 0; i < numCourses; i++) {
graph.add(new ArrayList<>());
}
for (int i = 0; i < prerequisites.length; i++) {
int cur = prerequisites[i][0];
int pre = prerequisites[i][1];
graph.get(pre).add(cur);
}
boolean[] visited = new boolean[numCourses];
for (int i = 0; i < numCourses; i++) {
if (!visited[i]) {
dfs(graph, i, visited, new boolean[numCourses], temp);
}
}
// 当最终序列中的课程数与总课程数不相同时，说明图中存在环
if (temp.size() != numCourses) {
return new int[0];
}
int[] ans = new int[numCourses];
Collections.reverse(temp);
for (int i = 0; i < numCourses; i++) {
ans[i] = temp.get(i);
}
return ans;
}
private void dfs(List<List<Integer>> graph, int x, boolean[] visited, boolean[] inStack, List<Integer> temp) {
visited[x] = true;
inStack[x] = true;
for (int next : graph.get(x)) {
// 若下一个将要遍历的课程还在递归栈中，说明存在环
if (inStack[next]) {
return;
}
if (!visited[next]) {
dfs(graph, next, visited, inStack, temp);
}
}
temp.add(x);
inStack[x] = false;
}
}
入度法
class Solution {
public int[] findOrder(int numCourses, int[][] prerequisites) {
List<Integer> temp = new ArrayList<>();
List<List<Integer>> graph = new ArrayList<>();
int[] inDegree = new int[numCourses];
Deque<Integer> q = new ArrayDeque<>();
// 生成图，注意图中的边是由前置课程指向当前课程
for (int i = 0; i < numCourses; i++) {
graph.add(new ArrayList<>());
}
for (int i = 0; i < prerequisites.length; i++) {
int cur = prerequisites[i][0];
int pre = prerequisites[i][1];
graph.get(pre).add(cur);
inDegree[cur]++;
}
for (int i = 0; i < numCourses; i++) {
if (inDegree[i] == 0) {
q.offer(i);
}
}
while (!q.isEmpty()) {
int x = q.poll();
for (int y : graph.get(x)) {
inDegree[y]--;
if (inDegree[y] == 0) {
q.offer(y);
}
}
temp.add(x);
}
// 当最终序列中的课程数与总课程数不相同时，说明图中存在环
if (temp.size() != numCourses) {
return new int[0];
}
int[] ans = new int[numCourses];
for (int i = 0; i < numCourses; i++) {
ans[i] = temp.get(i);
}
return ans;
}
}
JavaScript
postOrder
/**
* @param {number} numCourses
* @param {number[][]} prerequisites
* @return {number[]}
*/
var findOrder = function (numCourses, prerequisites) {
let ans = []
let visited = []
let edges = new Map(new Array(numCourses).fill(0).map((v, index) => [index, []]))
for (let edge of prerequisites) {
if (!edges.has(edge[1])) {
edges.set(edge[1], [])
}
edges.get(edge[1]).push(edge[0])
}
for (let i = 0; i < numCourses; i++) {
if (!visited[i]) {
dfs(i, edges, visited, [], ans)
}
}
return ans.length === numCourses ? ans.reverse() : []
}
let dfs = function (u, edges, visited, instack, ans) {
visited[u] = true
instack[u] = true
for (let v of edges.get(u)) {
if (instack[v]) {
return
}
if (!visited[v]) {
dfs(v, edges, visited, instack, ans)
}
}
ans.push(u)
instack[u] = false
}
入度法
/**
* @param {number} numCourses
* @param {number[][]} prerequisites
* @return {number[]}
*/
var findOrder = function (numCourses, prerequisites) {
let ans = []
let inDegree = new Array(numCourses).fill(0)
let edges = new Map(new Array(numCourses).fill(0).map((v, index) => [index, []]))
let q = []
for (let edge of prerequisites) {
inDegree[edge[0]]++
edges.get(edge[1]).push(edge[0])
}
for (let i = 0; i < numCourses; i++) {
if (!inDegree[i]) {
q.push(i)
}
}
while (q.length) {
let u = q.shift()
for (let v of edges.get(u)) {
inDegree[v]--
if (!inDegree[v]) {
q.push(v)
}
}
ans.push(u)
}
return ans.length === numCourses ? ans : []
}
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