[LeetCode] 207. Course Schedule 課程清單

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?

Example 1:

Input: 2, [[1,0]]
Output: true
Explanation: There are a total of 2 courses to take.
To take course 1 you should have finished course 0. So it is possible.

Example 2:

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

Hints:

1. 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.
2. There are several ways to represent a graph. For example, the input prerequisites is a graph represented by a list of edges. Is this graph representation appropriate?
3. Topological Sort via DFS - A great video tutorial (21 minutes) on Coursera explaining the basic concepts of Topological Sort.
4. Topological sort could also be done via BFS.

這道課程清單的問(wèn)題對(duì)于我們學(xué)生來(lái)說(shuō)應(yīng)該不陌生，因?yàn)樵谶x課的時(shí)候經(jīng)常會(huì)遇到想選某一門課程，發(fā)現(xiàn)選它之前必須先上了哪些課程，這道題給了很多提示，第一條就告訴了這道題的本質(zhì)就是在有向圖中檢測(cè)環(huán)。 LeetCode 中關(guān)于圖的題很少，有向圖的僅此一道，還有一道關(guān)于無(wú)向圖的題是 Clone Graph。個(gè)人認(rèn)為圖這種數(shù)據(jù)結(jié)構(gòu)相比于樹(shù)啊，鏈表啊什么的要更為復(fù)雜一些，尤其是有向圖，很麻煩。第二條提示是在講如何來(lái)表示一個(gè)有向圖，可以用邊來(lái)表示，邊是由兩個(gè)端點(diǎn)組成的，用兩個(gè)點(diǎn)來(lái)表示邊。第三第四條提示揭示了此題有兩種解法，DFS 和 BFS 都可以解此題。先來(lái)看 BFS 的解法，定義二維數(shù)組 graph 來(lái)表示這個(gè)有向圖，一維數(shù)組 in 來(lái)表示每個(gè)頂點(diǎn)的入度。開(kāi)始先根據(jù)輸入來(lái)建立這個(gè)有向圖，并將入度數(shù)組也初始化好。然后定義一個(gè) queue 變量，將所有入度為0的點(diǎn)放入隊(duì)列中，然后開(kāi)始遍歷隊(duì)列，從 graph 里遍歷其連接的點(diǎn)，每到達(dá)一個(gè)新節(jié)點(diǎn)，將其入度減一，如果此時(shí)該點(diǎn)入度為0，則放入隊(duì)列末尾。直到遍歷完隊(duì)列中所有的值，若此時(shí)還有節(jié)點(diǎn)的入度不為0，則說(shuō)明環(huán)存在，返回 false，反之則返回 true。代碼如下：

解法一：

class Solution {
public:
    bool canFinish(int numCourses, vector<vector<int>>& prerequisites) {
        vector<vector<int>> graph(numCourses, vector<int>());
        vector<int> in(numCourses);
        for (auto a : prerequisites) {
            graph[a[1]].push_back(a[0]);
            ++in[a[0]];
        }
        queue<int> q;
        for (int i = 0; i < numCourses; ++i) {
            if (in[i] == 0) q.push(i);
        }
        while (!q.empty()) {
            int t = q.front(); q.pop();
            for (auto a : graph[t]) {
                --in[a];
                if (in[a] == 0) q.push(a);
            }
        }
        for (int i = 0; i < numCourses; ++i) {
            if (in[i] != 0) return false;
        }
        return true;
    }
};

下面來(lái)看 DFS 的解法，也需要建立有向圖，還是用二維數(shù)組來(lái)建立，和 BFS 不同的是，像現(xiàn)在需要一個(gè)一維數(shù)組 visit 來(lái)記錄訪問(wèn)狀態(tài)，這里有三種狀態(tài)，0表示還未訪問(wèn)過(guò)，1表示已經(jīng)訪問(wèn)了，-1 表示有沖突。大體思路是，先建立好有向圖，然后從第一個(gè)門課開(kāi)始，找其可構(gòu)成哪門課，暫時(shí)將當(dāng)前課程標(biāo)記為已訪問(wèn)，然后對(duì)新得到的課程調(diào)用 DFS 遞歸，直到出現(xiàn)新的課程已經(jīng)訪問(wèn)過(guò)了，則返回 false，沒(méi)有沖突的話返回 true，然后把標(biāo)記為已訪問(wèn)的課程改為未訪問(wèn)。代碼如下：

解法二：

class Solution {
public:
    bool canFinish(int numCourses, vector<vector<int>>& prerequisites) {
        vector<vector<int>> graph(numCourses, vector<int>());
        vector<int> visit(numCourses);
        for (auto a : prerequisites) {
            graph[a[1]].push_back(a[0]);
        }
        for (int i = 0; i < numCourses; ++i) {
            if (!canFinishDFS(graph, visit, i)) return false;
        }
        return true;
    }
    bool canFinishDFS(vector<vector<int>>& graph, vector<int>& visit, int i) {
        if (visit[i] == -1) return false;
        if (visit[i] == 1) return true;
        visit[i] = -1;
        for (auto a : graph[i]) {
            if (!canFinishDFS(graph, visit, a)) return false;
        }
        visit[i] = 1;
        return true;
    }
};

Github 同步地址：

https://github.com/grandyang/leetcode/issues/207

參考資料：

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

https://leetcode.com/problems/course-schedule/discuss/58524/Java-DFS-and-BFS-solution

https://leetcode.com/problems/course-schedule/discuss/58516/Easy-BFS-Topological-sort-Java

https://leetcode.com/problems/course-schedule/discuss/162743/JavaC%2B%2BPython-BFS-Topological-Sorting-O(N-%2B-E)

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