Course Schedule II

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题目描述：

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].

解题思路：使用拓扑排序，使用一个数组保存每个课程前置课程的数目，使用一个数组保存每个科目的后置课程的编号，然后依次查找前置课程数目为0的课程输出，并使其所有后置课程的前置课程数目减1

AC代码如下（时间：144ms）

class Solution {
public:
     vector<int> findOrder(int numCourses, vector<pair<int, int>>& prerequisites) {
		vector<int> ans;
		if (numCourses == 0) return ans;
		vector<int> preCourseCount(numCourses, 0);
		vector<bool> processed(numCourses, false);
		vector<vector<int>> postCourseList(numCourses);
		int n = prerequisites.size();
		if (n == 0){
			for (int i = 0; i < numCourses; ++i){
				ans.push_back(i);
			}
			return ans;
		}
		for (int i = 0; i < n; ++i){
			int post = prerequisites[i].first;
			int pre = prerequisites[i].second;
			preCourseCount[post]++;
			postCourseList[pre].push_back(post);
		}
		while (1){
			if (ans.size() == numCourses) break;
			int i = 0;
			for (; i < numCourses; ++i){
				if (preCourseCount[i] == 0 && !processed[i]){
					processed[i] = true;
					ans.push_back(i);
					vector<int> temp = postCourseList[i];
					for (int j = 0; j < temp.size(); ++j){
						preCourseCount[temp[j]]--;
					}
					break;
				}
			}
			if (i >= numCourses) return vector<int>(0);
		}
		return ans;
	}
};

Course Schedule II

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原文地址：http://blog.csdn.net/lyh642784803/article/details/51345732