LeetCode: 207. Course Schedule

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LeetCode: 207. Course Schedule

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, is it possible for you to finish all courses?

Example 1:

Input: 2, [[1,0]] 
Output: true
Explanation: There are a total of 2 courses to take. 
             To take course 1 you should have finished course 0. So it is possible.

Example 2:

Input: 2, [[1,0],[0,1]]
Output: false
Explanation: 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.

Note:

1. The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.
2. You may assume that there are no duplicate edges in the input prerequisites.

解题思路 —— 拓扑排序

遍历所有节点找出没有入度的节点，然后删掉。重复操作，直到无法操作。如果最后有剩余的节点，则返回 false， 否则返回 true。

AC 代码

class Solution {
public:
    bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
        unordered_set<int> travelVertices;
        multimap<int, int> adjVer; // 邻接矩阵
        bool hasNewVer = true;
        
        for(auto iter = prerequisites.begin(); iter != prerequisites.end(); ++iter)
        {
            adjVer.insert({iter->second, iter->first});
        }
        
        while(travelVertices.size() != numCourses && hasNewVer)
        {
            // 记录每个课程的入度
            vector<int> courses(numCourses, 0);
            
            for(size_t i = 0; i < numCourses; ++i)
            {
                for(auto iter = adjVer.find(i); iter != adjVer.end() && iter->first == i; ++iter)
                {
                    if(travelVertices.find(iter->second) == travelVertices.end()) 
                    {
                    	++courses[iter->first];
                   }
                }
            }
            
            hasNewVer = false;
            for(size_t i = 0; i < courses.size(); ++i)
            {
                // 入度是 0 的课程可以上
                if(courses[i] == 0 && travelVertices.find(i) == travelVertices.end())
                {
                    travelVertices.insert(i);
                    hasNewVer = true;
                }
            }
        }
        
        return (travelVertices.size() == numCourses);
    }
};