版权声明：我的GitHub：https://github.com/617076674。真诚求星！ https://blog.csdn.net/qq_41231926/article/details/83510893
我的LeetCode代码仓：https://github.com/617076674/LeetCode
原题链接：https://leetcode-cn.com/problems/course-schedule-ii/description/
题目描述：
知识点：拓扑排序
思路：拓扑排序
本题的实现和LeetCode207——课程表一模一样，只不过在拓扑排序的时候多用了一个数组来记录顺序。
时间复杂度和空间复杂度均是O(n)。
JAVA代码：
public class Solution {
public int[] findOrder(int numCourses, int[][] prerequisites) {
int[] inDegree = new int[numCourses];
int[][] graph = new int[numCourses][numCourses];
for(int i = 0; i < prerequisites.length; i++){
graph[prerequisites[i][1]][prerequisites[i][0]] = 1;
inDegree[prerequisites[i][0]]++;
}
Queue<Integer> queue = new LinkedList<>();
for(int i = 0; i < numCourses; i++){
if(inDegree[i] == 0){
queue.add(i);
}
}
int[] result = new int[numCourses];
int index = 0;
int num = 0;
while(!queue.isEmpty()){
int u = queue.poll();
result[index++] = u;
for(int v = 0; v < numCourses; v++){
if(graph[u][v] != 0){
inDegree[v]--;
if(inDegree[v] == 0){
queue.add(v);
}
}
}
num++;
}
if(num != numCourses){
result = new int[0];
}
return result;
}
}JAVA解题报告：
C++代码：
class Solution {
public:
vector<int> findOrder(int numCourses, vector<pair<int, int>>& prerequisites) {
int inDegree[numCourses] = {0};
vector<int> graph[numCourses];
for(int i = 0; i < prerequisites.size(); i++){
graph[prerequisites[i].second].push_back(prerequisites[i].first);
inDegree[prerequisites[i].first]++;
}
queue<int> q;
for(int i = 0; i < numCourses; i++){
if(inDegree[i] == 0){
q.push(i);
}
}
int num = 0;
vector<int> result;
while(!q.empty()){
int u = q.front();
result.push_back(u);
q.pop();
for(int j = 0; j < graph[u].size(); j++){
int v = graph[u][j];
inDegree[v]--;
if(inDegree[v] == 0){
q.push(v);
}
}
num++;
}
if(num != numCourses){
result.clear();
}
return result;
}
};C++解题报告：
