[LeetCode] 207. Course Schedule

題目鏈接: https://leetcode.com/problems/course-schedule/description/

Description

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, is it possible for you to finish all courses?

For example:

2, [[1,0]]

There are a total of 2 courses to take. To take course 1 you should have finished course 0. So it is possible.

2, [[1,0],[0,1]]

There are a total of 2 courses to take. To take course 1 you should have finished course 0, and to take course 0 you should also have finished course 1. So it is impossible.

Note:

1. The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.
2. You may assume that there are no duplicate edges in the input prerequisites.

解題思路

題意爲給定課程數 n 和先導課程數組，數組裏每一個元素爲 [課程號, 先導課程號]，要學習某門課程，需要先學習先導課程，問是否可以完成所有課程。

題目可以轉化爲有向圖，每一個課程爲一個頂點，有向邊由 先導課程 指向 目標課程，判斷該有向圖是否有環，無環則可以完成所有課程，否則不能完成。

判斷一個有向圖是否有環有很多方法，如 DFS、BFS、拓撲排序。這裏使用的是拓撲排序的方法，爲方便找每個頂點可以到達的頂點，用了一個 unordered_map 將先導課程數組轉化爲鄰接鏈表形式，也可以用 vector 替代。記錄每一個頂點的入度，然後就跟拓撲排序一樣了。最後再判斷一下入度數組是否有元素不爲 0，若有則有環，返回 fasle，否則返回 true。

Code

class Solution {
public:
    bool canFinish(int numCourses, vector<pair<int, int>> &prerequisites) {
        vector<int> indegree(numCourses, 0);
        unordered_map<int, list<int>> umap;
        queue<int> mq;

        for (int i = 0; i < prerequisites.size(); ++i) {
            if (umap.count(prerequisites[i].second) == 0)
                umap[prerequisites[i].second] = list<int>(1, prerequisites[i].first);
            else
                umap[prerequisites[i].second].push_back(prerequisites[i].first);
            indegree[prerequisites[i].first]++;
        }

        for (int i = 0; i < numCourses; ++i) {
            if (indegree[i] == 0) mq.push(i);
        }

        while (!mq.empty()) {
            int start = mq.front();
            mq.pop();

            for (auto it = umap[start].begin(); it != umap[start].end(); ++it) {
                indegree[*it]--;
                if (indegree[*it] == 0) mq.push(*it);
            }
        }

        for (auto x: indegree) if (x != 0) return false;
        return true;
    }
};