Leetcode: Course Schedule

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:
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.
click to show more hints.

Hints:
This problem is equivalent to finding if a cycle exists 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.

問題描述及算法分析

課程清單這題是典型的拓撲排序，本題Note中也提到其實質就是判斷在一個有向圖中是否有環。拓撲排序有兩個基本操作：決定一個頂點是否入度爲0和刪除一個頂點的所有出邊。即在有向圖中每次找到一個入度爲0的節點，然後把它指向其他節點的邊都去掉，重複這個過程，直到所有節點已被找到，或者沒有符合條件的節點（有環存在）。
算法採用隊列存儲節點可降低複雜度，二維數組vector<vector<int>>表示一個圖，用一維數組vector<int>記錄入度的個數，定義一個queue變量，將所有入度爲0的點放入隊列中，然後開始遍歷隊列，從graph裏遍歷其連接的點，每到達一個新節點，將其入度減一，如果此時該點入度爲0，則放入隊列末尾。直到遍歷完隊列中所有的值，若此時還有節點的入度不爲0，則說明環存在，返回false，反之則返回true。

代碼

class Solution {
public:
    bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
        vector<vector<int>> graph(numCourses);
        vector<int> indegree(numCourses,0);
        for(auto i : prerequisites) {
            graph[i.first].push_back(i.second);
            indegree[i.second]++;
        }
        queue<int> q;
        for (auto a = 0; a < numCourses;a++) {
            if (indegree[a] == 0) {
                q.push(a);
            }
        }
        int count = 0;
        while (!q.empty()) {
            int temp = q.front();
            q.pop();
            count++;
            for(auto v : graph[temp]) {
                indegree[v]--;
                if (indegree[v] == 0) {
                    q.push(v);
                }
            }
        }
        return count == numCourses;
    }
};

【易錯點】vector容器的賦值及遍歷操作；隊列的聲明、入隊及出隊操作

其他算法分析

可採用DFS算法實現

class Solution {
public:
    bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
        vector<vector<int>> graph(numCourses);
        vector<int> visited(numCourses, 0);
        for (auto a : prerequisites) {
            graph[a.second].push_back(a.first);
        }
        bool cycle = false;
        for (auto b = 0; b < numCourses; b++) {
            if (cycle) return false;
            if (visited[b] == 0) {
                DFS(b,graph,cycle,visited);
            }
        }
        return !cycle;
    }
    void DFS(int node, vector<vector<int>>& graph, bool & cycle, vector<int> & visited) {
        if(visited[node] == 1) {
            cycle = true;
            return;
        }
        visited[node] = 1;
        for (auto i : graph[node]) {
            DFS(i,graph,cycle,visited);
            if(cycle) return;
        }
        visited[node] = 2;
    }
};