LeetCode 207. Course Schedule

LeetCode 207. Course Schedule：题目、思路与代码实现。

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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本文最初发布于简书。

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