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:
- 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.
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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