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[LeetCode](week3) 207. Course Sc

[LeetCode](week3) 207. Course Sc

作者: jeff98 | 来源:发表于2018-09-23 16:22 被阅读0次

    这周学了图,来做一下这题

    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.

    题目分析

    一开始的思路:访问时pre未访问过?

    哪个方向建图

    解析为:判断图中有无环

    别忘了内存

    题解

    #include <iostream>
    #include <vector>
    
    using namespace std;
    class Solution {
    public:
        bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
    
            vector<int>* table = new vector<int>[numCourses];
            int* inDegree = new int[numCourses];
            for(int i = 0; i < numCourses; ++i)
                inDegree[i] = 0;
            
            for(vector<pair<int, int>>::iterator iter = prerequisites.begin(); iter != prerequisites.end(); iter++){
                inDegree[iter->second]++;
                table[iter->first].push_back(iter->second);
            }
    
            for(int i = 0; i < numCourses; ++i){
                if(inDegree[i] == 0){
                    //decrease the inDegree of children
                    for(vector<int>::iterator iter = table[i].begin(); iter != table[i].end(); iter++){
                        inDegree[*iter] --;
                    }
                    //remove : set inDegree to -1
                    inDegree[i] = -1;
                    i = -1; //back from top to scan
                }
            }
    
            //check for unremoved(inDegree != -1)
            for(int i = 0; i < numCourses; ++i){
                if(inDegree[i] != -1)
                    return false;
            }
            //memory?!!
            return true;
    
        }
    };
    

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