leetcode - Course Schedule II

标签：c++   程序员   面试   拓扑排序   

题目：leetcode

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. Read more about how a graph is represented.

click to show more hints.

分析：

运用了拓扑排序，参考这里。

一个简单的拓扑排序算法：

a）先找到一个没有输入边的点，输出这个点，然后去掉与这个点连接的所有边。

b）重复上面的步骤知道输出所有的点。

class Solution {
public:
    vector<int> findOrder(int numCourses, vector<pair<int, int>>& prerequisites) {
        //输入居然有重复值！！！
        sort(prerequisites.begin(),prerequisites.end());
        prerequisites.erase(unique(prerequisites.begin(),prerequisites.end()),prerequisites.end());
        
         vector<int> res;
         if(numCourses==0 )
           return res;
         if(prerequisites.empty())
        {
            for(int i=0;i<numCourses;++i)
            res.push_back(i);
            return res;
        }
        
        unordered_map<int,vector<int>> end_start;
        for(auto &i:prerequisites)
        {
            end_start[i.first].push_back(i.second);
        }
        unordered_set<int> num;
        for(int i=0;i<numCourses;++i)
            num.emplace(i);
        while(!num.empty())
        {
            bool should_return=true;
            for(auto &n:num)
            {
                if(end_start.find(n)==end_start.end())
                {
                    should_return=false;
                    res.push_back(n);
                    cache_erase(end_start,n);
                    num.erase(n);
                    break;
                }
            }
            if(should_return)
                return vector<int>();
        }
        
       
        return res;
    }
    
    void cache_erase(unordered_map<int,vector<int>> &end_start,int n)
    {
        for(auto it=end_start.begin();it!=end_start.end();)
        {
            auto tmp=find(it->second.begin(),it->second.end(),n);
            if(tmp!=it->second.end())
            {
                it->second.erase(tmp);
                if(it->second.empty())
                {
                    end_start.erase(it++);
                }
                else
                    ++it;
            }
            else
                ++it;
        }
    }
    
    
};

leetcode - Course Schedule II

标签：c++   程序员   面试   拓扑排序   

原文地址：http://blog.csdn.net/bupt8846/article/details/45717669