Course Schedule题解

原创文章,拒绝转载

题目来源:https://leetcode.com/problems/course-schedule/description/


Description

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?

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:

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.

Solution

class Solution {
private:
bool **graph;
int* color;
int graphDim = 0; public:
~Solution() {
if (graphDim == 0)
return;
delete[] color;
for (int i = 0; i < graphDim; i++)
delete[] graph[i];
delete graph;
} bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
if (numCourses <= 0)
return false;
if (prerequisites.size() == 0)
return true;
int i, j; graphDim = numCourses;
graph = new bool*[numCourses];
color = new int[numCourses]; for (i = 0; i < numCourses; i++)
graph[i] = new bool[numCourses]; for (i = 0; i < numCourses; i++) {
color[i] = 0;
for (j = 0; j < numCourses; j++) {
graph[i][j] = false;
}
} for (auto p: prerequisites) {
graph[p.first][p.second] = true;
} int sv;
for (auto p: prerequisites) {
sv = p.first;
for (i = 0; i < numCourses; i++) {
if (graph[sv][i]) {
if (color[sv] == 2)
continue;
if (!dfs(sv))
return false;
else
color[i] = 2;
}
}
}
return true;
} bool dfs(int v) {
color[v] = 1;
int i;
for (i = 0; i < graphDim; i++) {
if (graph[v][i]) {
if (color[i] == 2) {
break;
} else if (color[i] == 1) {
return false;
} else {
if (!dfs(i)) {
return false;
}
}
}
}
color[v] = 2;
return true;
}
};

解题描述

这道题我使用的是DFS,过程中通过对以访问的顶点进行染色来判断是否有环。不过提交后出现了几次TLE,检查代码发现好几处可以充分利用染色数组来减少充分访问。改了好几次才AC。现实利用中DFS和BFS处理的数据量往往会很大,这几次TLE还是提醒我要注意优化算法时间复杂度。


文章更新

这道题本质上是检查图是否有环,而如果图是可以拓扑排序的,则一定是无环的。通过BFS检查图是否能够拓扑排序也是可以解决的,且相对来说实现效率较高,避免了DFS的递归开销。下面给出实现:

class Solution
{
private:
map<int, int> inDegree;
public:
bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
int i, curVertex;
for (i = 0; i < numCourses; i++)
inDegree[i] = 0; for (auto p: prerequisites)
inDegree[p.second]++; queue<int> vq;
for (i = 0; i < numCourses; i++) {
if (inDegree[i] == 0) {
vq.push(i);
}
} if (vq.empty())
return false; while (!vq.empty()) {
curVertex = vq.front();
vq.pop();
inDegree.erase(curVertex);
for (i = 0; i < prerequisites.size(); i++) {
if (curVertex == prerequisites[i].first) {
inDegree[prerequisites[i].second]--;
if (inDegree[prerequisites[i].second] == 0)
vq.push(prerequisites[i].second);
}
}
} return inDegree.empty();
}
};

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