LeetCode 210. Course Schedule II（拓撲排序）

LeetCode 210. Course Schedule II

Medium

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.

題意

給定有向圖的邊清單，輸出拓撲排序的一個可行結果。如果有向圖有環，則輸出空。

思路

拓撲排序，和LeetCode 207. Course Schedule基本一樣。可以題目中要求有向邊(from_id, to_id)中to_id在前，from_id在後，是以用出度為0可以直接求解得要求的拓撲序，用入度為0求解的拓撲序要reverse才符合題目要求。

代碼

• 出度為0版本

class Solution {
public:
    vector<int> findOrder(int numCourses, vector<vector<int>>& prerequisites) {
        vector<vector<int>> adj(numCourses, vector<int>()); // adjacent list from from_id to to_id_set
        vector<int> outdegree(numCourses, 0);   // in-degree
        for (auto edge: prerequisites) {
            adj[edge[1]].push_back(edge[0]);
            outdegree[edge[0]]++;
        }
        int sum = 0;        // sum of nodes in topoSort
        vector<int> zerodegree; // stack of zero degree nodes
        vector<int> topoOrder;  // topo-order array
        for (int i=0; i<numCourses; ++i) {
            if (outdegree[i] == 0) {
                zerodegree.push_back(i);
                ++sum;
            }
        }
        while (!zerodegree.empty()) {   // topoSort
            int i = zerodegree.back();
            topoOrder.push_back(i);
            zerodegree.pop_back();
            for (auto v: adj[i]) {
                if (!(--outdegree[v])) {
                    zerodegree.push_back(v);
                    ++sum;
                }
            }
        }
        return sum == numCourses? topoOrder: vector<int>();
    }
};
           

• 入度為0版本

class Solution {
public:
    vector<int> findOrder(int numCourses, vector<vector<int>>& prerequisites) {
        vector<vector<int>> adj(numCourses, vector<int>()); // adjacent list from from_id to to_id_set
        vector<int> indegree(numCourses, 0);   // in-degree
        for (auto edge: prerequisites) {
            adj[edge[0]].push_back(edge[1]);
            indegree[edge[1]]++;
        }
        int sum = 0;        // sum of nodes in topoSort
        vector<int> zerodegree; // stack of zero degree nodes
        vector<int> topoOrder;  // topo-order array
        for (int i=0; i<numCourses; ++i) {
            if (indegree[i] == 0) {
                zerodegree.push_back(i);
                ++sum;
            }
        }
        while (!zerodegree.empty()) {   // topoSort
            int i = zerodegree.back();
            topoOrder.push_back(i);
            zerodegree.pop_back();
            for (auto v: adj[i]) {
                if (!(--indegree[v])) {
                    zerodegree.push_back(v);
                    ++sum;
                }
            }
        }
        reverse(topoOrder.begin(), topoOrder.end());
        return sum == numCourses? topoOrder: vector<int>();
    }
};