210 Course Schedule II

Author: int32bit

README.md

@@ -116,6 +116,7 @@
 + [206 Reverse Linked List(链表转置，递归&迭代)](algorithms/ReverseLinkedList)
 + [207 Course Schedule(拓扑排序、图是否存在环、BFS、DFS)](algorithms/CourseSchedule)
 + [208 Implement Trie (Trie树、字典树、前缀树)](algorithms/ImplementTrie)
++ [210 Course Schedule II(拓扑排序)](algorithms/CourseScheduleII)
 + [211 Add and Search Word - Data structure design(Trie树)](algorithms/AddandSearchWord)
 
 ## Database

algorithms/CourseScheduleII/README.md

@@ -0,0 +1,59 @@
+## Course Schedule II
+
+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.
+
+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 the correct course order is `[0,1]`
+
+```
+4, [[1,0],[2,0],[3,1],[3,2]]
+```
+
+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.*
+
+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.
+
+## Solution
+
+求拓扑排序结果, 算法见[Course Schedule](../Course Schedule)
+
+```cpp
+vector<int> topsort(int n, const vector<pair<int, int>> &request) {
+	vector<int> result;
+	vector<int> degree(n, 0);
+	for (auto p : request) {
+		degree[p.first]++;
+	}
+	while (result.size() < n) { // 还没有访问完
+		int cur = findZero(degree);
+		if (cur >= 0) { // 访问节点cur
+			result.push_back(cur);
+			degree[cur] = -1; // 标记当前节点为已经访问状态
+			for (auto p : request) {
+				if (p.second == cur)
+					degree[p.first]--; // 去掉已访问节点
+			}
+		} else
+			return vector<int>();
+	}
+	return result;
+}
+```

algorithms/CourseScheduleII/bfs.cpp

@@ -0,0 +1,55 @@
+#include <cstdio>
+#include <cstdlib>
+#include <vector>
+#include <algorithm>
+#include <utility>
+#include <iostream>
+using namespace std;
+class Solution {
+	public:
+		vector<int> findOrder(int n, vector<pair<int, int>> &request) {
+			return topsort(n, request);
+		}
+	private:
+		/* BFS */
+		vector<int> topsort(int n, const vector<pair<int, int>> &request) {
+			vector<int> result;
+			vector<int> degree(n, 0);
+			for (auto p : request) {
+				degree[p.first]++;
+			}
+			while (result.size() < n) { // 还没有访问完
+				int cur = findZero(degree);
+				if (cur >= 0) { // 访问节点cur
+					result.push_back(cur);
+					degree[cur] = -1; // 标记当前节点为已经访问状态
+					for (auto p : request) {
+						if (p.second == cur)
+							degree[p.first]--; // 去掉已访问节点
+					}
+				} else
+					return vector<int>();
+			}
+			return result;
+		}
+		int findZero(const vector<int> &v) {
+			int n = v.size();
+			for (int i = 0; i < n; ++i) {
+				if (v[i] == 0)
+					return i;
+			}
+			return -1;
+		}
+
+};
+void print(vector<int> v) {
+	for_each(begin(v), end(v), [](int i){cout << i << ' ';});
+	cout << endl;
+}
+int main(int argc, char **argv)
+{
+	Solution solution;
+	vector<pair<int, int>> request = {make_pair(1, 0), make_pair(2, 0), make_pair(3, 1), make_pair(3, 2)};
+	print(solution.findOrder(4, request));
+	return 0;
+}