210. Course Schedule II（课程清单之二）

这道题没有什么好说的，因为基本解法都在上一篇博客（207. Course Schedule），这题就是多加一个vector来记录路径而已。

BFS做法：

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

DFS做法：

bool DFS(vector<vector<int>>& graph, vector<int>& visit, int i, vector<int>& path) {
	if (visit[i] == -1)
		return false;
	if (visit[i] == 1)
		return true;
	visit[i] = -1;//等于-1代表不可重复
	for (auto t : graph[i]) {
		if (!DFS(graph, visit, t, path))
			return false;
	}
	visit[i] = 1;//等于1代表已访问
    path.push_back(i);
	return true;
}

class Solution {
public:
    vector<int> findOrder(int numCourses, vector<pair<int, int>>& prerequisites) {
        vector<int> visit(numCourses, 0);
        vector<vector<int>> graph(numCourses, vector<int>());
        vector<int> path;

        for (auto t : prerequisites) {
            graph[t.first].push_back(t.second);
        }

        for (int i = 0; i < graph.size(); i++) {
            if (!DFS(graph, visit, i, path))
                return vector<int>();
        }
        return path;
    }
};