[LeetCode] 210. Course Schedule II 課程清單之二

時間 2019-11-07

標籤 leetcode course schedule 課程清單之二简体版

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There are a total of n courses you have to take, labeled from 0 to n-1.html

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]git

Given the total number of courses and a list of prerequisite pairs, return the ordering of courses you should take to finish all courses.github

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.express

Example 1:數組

Input: 2, [[1,0]] 
Output: 
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][0,1] .

Example 2:ide

Input: 4, [[1,0],[2,0],[3,1],[3,2]]
Output: 
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 . Another correct ordering is [0,1,2,3] or [0,2,1,3][0,1,2,3][0,2,1,3] .

Note:post

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.

Hints:

This problem is equivalent to finding the topological order in a directed graph. If a cycle exists, no topological ordering exists and therefore it will be impossible to take all courses.
Topological Sort via DFS - A great video tutorial (21 minutes) on Coursera explaining the basic concepts of Topological Sort.
Topological sort could also be done via BFS.

這題是以前那道 Course Schedule 的擴展，那道題只讓咱們判斷是否能完成全部課程，即檢測有向圖中是否有環，而這道題咱們得找出要上的課程的順序，即有向圖的拓撲排序 Topological Sort，這樣一來，難度就增長了，可是因爲咱們有以前那道的基礎，而此題正是基於以前解法的基礎上稍加修改，咱們從 queue 中每取出一個數組就將其存在結果中，最終如有向圖中有環，則結果中元素的個數不等於總課程數，那咱們將結果清空便可。代碼以下：ui

class Solution {
public:
    vector<int> findOrder(int numCourses, vector<pair<int, int>>& prerequisites) {
        vector<int> res;
        vector<vector<int> > graph(numCourses, vector<int>(0));
        vector<int> in(numCourses, 0);
        for (auto &a : prerequisites) {
            graph[a.second].push_back(a.first);
            ++in[a.first];
        }
        queue<int> q;
        for (int i = 0; i < numCourses; ++i) {
            if (in[i] == 0) q.push(i);
        }
        while (!q.empty()) {
            int t = q.front();
            res.push_back(t);
            q.pop();
            for (auto &a : graph[t]) {
                --in[a];
                if (in[a] == 0) q.push(a);
            }
        }
        if (res.size() != numCourses) res.clear();
        return res;
    }
};