LeetCode刷题(C++)——Course Schedule
2017-08-16 20:52
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There are a total of n courses you have to take, labeled from
Some courses may have prerequisites, for example to take course 0 you have to first take course 1, which is expressed as a pair:
Given the total number of courses and a list of prerequisite pairs, is it possible for you to finish all courses?
For example:
There are a total of 2 courses to take. To take course 1 you should have finished course 0. So it is possible.
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:
The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how
a graph is represented.
You may assume that there are no duplicate edges in the input prerequisites.
class Solution {
public:
bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
if(prerequisites.empty())
return true;
vector<int> inDegree(numCourses);
for(auto x : prerequisites){
inDegree[x.first]++;
}
queue<int> q;
for (int i = 0; i < numCourses; ++i) {
if (inDegree[i] == 0)
q.push(i);
}
int count = 0;
while(!q.empty()){
int u = q.front();
q.pop();
count++;
for(auto x : prerequisites){
if(x.second == u){
inDegree[x.first]--;
if(inDegree[x.first]==0)
q.push(x.first);
}
}
}
if(count==numCourses)
return true;
return false;
}
};
0to
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.
Note:
The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how
a graph is represented.
You may assume that there are no duplicate edges in the input prerequisites.
class Solution {
public:
bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
if(prerequisites.empty())
return true;
vector<int> inDegree(numCourses);
for(auto x : prerequisites){
inDegree[x.first]++;
}
queue<int> q;
for (int i = 0; i < numCourses; ++i) {
if (inDegree[i] == 0)
q.push(i);
}
int count = 0;
while(!q.empty()){
int u = q.front();
q.pop();
count++;
for(auto x : prerequisites){
if(x.second == u){
inDegree[x.first]--;
if(inDegree[x.first]==0)
q.push(x.first);
}
}
}
if(count==numCourses)
return true;
return false;
}
};
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