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C++實(shí)現(xiàn)LeetCode(210.課程清單之二)

更新時(shí)間：2021年08月10日 15:12:55 作者：Grandyang

這篇文章主要介紹了C++實(shí)現(xiàn)LeetCode(210.課程清單之二),本篇文章通過(guò)簡(jiǎn)要的案例,講解了該項(xiàng)技術(shù)的了解與使用,以下就是詳細(xì)內(nèi)容,需要的朋友可以參考下

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

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.

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.

這題是之前那道 Course Schedule 的擴(kuò)展，那道題只讓我們判斷是否能完成所有課程，即檢測(cè)有向圖中是否有環(huán)，而這道題我們得找出要上的課程的順序，即有向圖的拓?fù)渑判?Topological Sort，這樣一來(lái)，難度就增加了，但是由于我們有之前那道的基礎(chǔ)，而此題正是基于之前解法的基礎(chǔ)上稍加修改，我們從 queue 中每取出一個(gè)數(shù)組就將其存在結(jié)果中，最終若有向圖中有環(huán)，則結(jié)果中元素的個(gè)數(shù)不等于總課程數(shù)，那我們將結(jié)果清空即可。代碼如下：

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;
    }
};

Github 同步地址：

https://github.com/grandyang/leetcode/issues/210

類(lèi)似題目：

Course Schedule

參考資料：

https://leetcode.com/problems/course-schedule-ii/

https://leetcode.com/problems/course-schedule-ii/discuss/59330/Concise-JAVA-solution-based-on-BFS-with-comments

https://leetcode.com/problems/course-schedule-ii/discuss/59342/Java-DFS-double-cache-visiting-each-vertex-once-433ms

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