Leetcode 207. Course Schedule

Leetcode 207. Course Schedule

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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?

输入：课程数量，以及各课程前序课程的pair

输出：是否能完成课程

如：

Input:
2, [[1,0]]
2, [[1,0],[0,1]]

Output:
true
false

Explain:
上述输入2中，两个课程互为前序课程，无法完成

思路

简单来说，这是判断有向图中是否产生环路。这里选择使用拓扑排序的方式来检查，如果无法完成拓扑排序，也就是该课程有向图中产生了环路，即无法完成。

选用Kahn算法。

L← Empty list that will contain the sorted elements
S ← Set of all nodes with no incoming edges
while S is non-empty do
remove a node n from S
insert n into L
foreach node m with an edge e from nto m do
remove edge e from thegraph
ifm has no other incoming edges then
insert m into S
if graph has edges then
return error (graph has at least onecycle)
else
return L (a topologically sortedorder)

代码

class Solution {
public:
set<int> s;

bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
int comingEdge[numCourses] = {0};
list<int> graph[numCourses];
int n,m;//node: n-e->m
//build the list and counting comingEdge
for(int i=0;i<prerequisites.size();i++){
n = prerequisites[i].second;
m = prerequisites[i].first;
//cout<<"m=="<<m<<" n=="<<n<<endl;
comingEdge[m]++;
graph
.push_back(m);
}
//init Set S
for(int i =0;i<numCourses;i++){
if(comingEdge[i]==0)
s.insert(i);
}
while(!s.empty()){
int nodeN = *s.begin();
s.erase(nodeN);
maintainSet(nodeN,graph[nodeN],comingEdge);
}

return isNoComingEdge(comingEdge,numCourses);
}

void maintainSet(int n, list<int>& graph,int* comingEdge){
//cout<<n<<" pop:";
while(graph.size()!=0){
int m = graph.front();
//cout<<m;
graph.pop_front();
comingEdge[m]--;
if(comingEdge[m]==0){
s.insert(m);
}
}
//cout<<endl;

}
bool isNoComingEdge(int* comingEdge, int numCourses){
for(int i =0;i<numCourses;i++){
if(comingEdge[i]!=0){
//cout<<"i:"<<" "<<comingEdge[i]<<endl;
return false;
}
}
return true;
}

};