leetcode-Course Schedule

标签：c++   程序员   面试   回溯法   哈希表   

题目：leetcode

Course Schedule

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?

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 it is possible.

2, [[1,0],[0,1]]

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.

分析：

本题目可转化为 “判断有向图中是否有环”。

1、把数组prerequisites中的内容放进哈希表table，table的key是课程编号，value是一个数组，记录了key代表的课程的约束条件（即上完value中的课程，才能上key的课程）。如果每一门课程当成平面上的一个点，那么key和value中的每一门课程，可以分别连成一条有向路径。

2、利用回溯法，如果没有环，则把相应的“路径”删除，直到把有向图删光。若有环，则整个程序返回false。

class Solution {
public:
    bool canFinish(int numCourses, vector<vector<int>>& prerequisites) {
        if(prerequisites.size()<=1)
            return true;
        
        unordered_map<int,vector<int>> table;
        for(auto &i:prerequisites)
        {
            table[i[0]].push_back(i[1]);
        }
       vector<int> path;
        while(!table.empty())
        {
            auto it=table.begin();
            path.push_back(it->first);
            if(HasLoop(table,it->first,path))
                return false;
            path.pop_back();
        }
        return true;
    }
    //判断是否有环，有环<span style="font-family: Arial, Helvetica, sans-serif;">的话返回真，否则返回假</span>
    bool HasLoop( unordered_map<int,vector<int>> &table,const int &begin,vector<int> &path)
    {
       if(table.count(begin)==0)
         return false;
       while(!table[begin].empty())
       {
           int temp=table[begin].back();
           if(find(path.begin(),path.end(),temp)!=path.end())
                return true;
            path.push_back(temp);
           if(HasLoop(table,temp,path))
                return true;
            path.pop_back();
            table[begin].pop_back();
       }
       table.erase(begin);
       return false;
    }
    
};

leetcode-Course Schedule

标签：c++   程序员   面试   回溯法   哈希表   

原文地址：http://blog.csdn.net/bupt8846/article/details/45543865