There are a total of n courses you have to take, labeled from
0ton - 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.
大致意思就是給定我們一定數(shù)量的課程，以及課程間的先修關(guān)系，讓我們判斷是否可以修完所有課程。
該問題等價于判斷一個有向圖是否有環(huán)，如果有環(huán)，就無法對該有向圖進(jìn)行拓?fù)渑判?/a>，也就無法在滿足課程間的先修關(guān)系的前提下修完所有課程。
我們可以依次從每個頂點開始使用DFS進(jìn)行圖的搜索，一旦發(fā)現(xiàn)環(huán)即可斷定無法修完所有課程。為了提高效率，本算法加了一個回溯拆邊的操作，因為既然已經(jīng)確認(rèn)這條路徑上沒有環(huán)，就不需要重復(fù)訪問了。
代碼如下：
public class Solution {
public boolean canFinish(int numCourses, int[][] prerequisites) {
boolean[] visited = new boolean[numCourses]; // 訪問標(biāo)志
List<Integer>[] adj = new List[numCourses]; // 鄰接鏈表
for(int i = 0; i < numCourses; i++)
adj[i] = new ArrayList<Integer>();
for(int i = 0; i < prerequisites.length; i++) // 由偏序關(guān)系構(gòu)造鄰接鏈表
{
int curCourse = prerequisites[i][0];
int preCourse = prerequisites[i][1];
adj[preCourse].add(curCourse);
}
for(int i = 0; i < numCourses; i++) // 從每一個結(jié)為起點開始做DFS，每次循環(huán)visited數(shù)組元素均為false
{
if(!dfs(adj, visited, i)) // 任意一次DFS發(fā)現(xiàn)環(huán)均可斷定無法修完所有課程
return false;
}
return true;
}
private boolean dfs(List<Integer>[] adj, boolean[] visited, int course){
if(visited[course]) // 再次訪問訪問過的結(jié)點，有環(huán)
return false;
visited[course] = true;
for (int i = 0; i < adj[course].size(); i++)
{
if(!dfs(adj, visited, adj[course].get(i)))
return false;
adj[course].remove(i); // 未發(fā)現(xiàn)環(huán)，依次拆掉走過的邊，避免再走這條路
}
visited[course] = false; // 標(biāo)志復(fù)位，向上回溯
return true;
}
}
解決本題的另一種方法是使用BFS，以無需先修的課程為起點，根據(jù)課程間的偏序關(guān)系搜索其他課程，統(tǒng)計搜索到（修完）的課程數(shù)，與總課程數(shù)進(jìn)行比較即可判斷是否可以修完所有課程，代碼如下：
public class Solution {
public boolean canFinish(int numCourses, int[][] prerequisites) {
List<Integer>[] adj = new List[numCourses]; // 鄰接鏈表
for(int i = 0; i < numCourses; i++)
adj[i] = new ArrayList<Integer>();
int[] indegree = new int[numCourses]; // 入度，即所需先修課程數(shù)
Queue<Integer> readyCourses = new LinkedList(); // 可修課程隊列
int finishCount = 0; // 修完的課程數(shù)
for (int i = 0; i < prerequisites.length; i++) // 由偏序關(guān)系構(gòu)造鄰接鏈表，并統(tǒng)計入度
{
int curCourse = prerequisites[i][0];
int preCourse = prerequisites[i][1];
adj[preCourse].add(curCourse);
indegree[curCourse]++;
}
for (int i = 0; i < numCourses; i++)
{
if (indegree[i] == 0)
readyCourses.offer(i); // 無需先修的課程先入隊列
}
while (!readyCourses.isEmpty())
{
int course = readyCourses.poll(); // 出隊列，即修完該課程
finishCount++;
for (int nextCourse : adj[course])
{
indegree[nextCourse]--;
if (indegree[nextCourse] == 0) // 所有先修課程已修完，入隊列
readyCourses.offer(nextCourse);
}
}
return finishCount == numCourses;
}
}