拓撲排序例項
在前面的文章《》中介紹了拓撲排序，在《》中給出了一種Java實現，其實現方式是從出度為0的點（也就是它不依賴任何其他點）逆向遍歷的。其實還可以正向遍歷，過程在前一篇文章的虛擬碼中給出。這裡看兩道相關題目加深理解應用。
1.Course Schedule
There are a total of n courses you have to take, labeled from0 ton - 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 prerequisitepairs, 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 abouthow
a graph is represented.
下面給出遞迴與非遞迴版本
遞迴：
public class Solution {
boolean[] used;
boolean[] global;
// LinkedList<Integer> res = new LinkedList<Integer>();
public boolean canFinish(int numCourses, int[][] prerequisites) {
ArrayList[] graph = new ArrayList[numCourses];
for(int[] pre:prerequisites){
if(graph[pre[0]] == null) graph[pre[0]] = new ArrayList<Integer>();
graph[pre[0]].add(pre[1]);
}
used = new boolean[numCourses];
global = new boolean[numCourses];
for(int i = 0;i < numCourses;i++){
if(global[i]) continue;
Arrays.fill(used,false);
if(!dfs(graph,i)) return false;
}
return true;
}
public boolean dfs(List<Integer>[] graph,int x){
boolean res = true;
used[x] = true;
global[x] = true;
if(graph[x] == null) return true;
for(int y:graph[x]){
if(used[y]) return false;
// res.addFirst(y);
res = res & dfs(graph,y);
used[y] = false;
}
return res;
}
}非遞迴版本
public class Solution {
boolean[] used;
boolean[] global;
// LinkedList<Integer> res = new LinkedList<Integer>();
public boolean canFinish(int numCourses, int[][] prerequisites) {
ArrayList[] graph = new ArrayList[numCourses];
for(int[] pre:prerequisites){
if(graph[pre[0]] == null) graph[pre[0]] = new ArrayList<Integer>();
graph[pre[0]].add(pre[1]);
}
global = new boolean[numCourses];
used = new boolean[numCourses];
LinkedList<Integer> stack = new LinkedList<Integer>();
for(int i = 0;i < numCourses;i++){
Arrays.fill(used,false);
if(global[i]) continue;
stack.addLast(i);
used[i] = true;
global[i] = true;
if(graph[i] == null) {stack.removeLast();used[i] = false;continue;}
boolean flag = false;
while(stack.size() > 0){
int course = stack.getLast();
if(graph[course] == null){stack.removeLast(); used[course] = false;continue;}
flag = false;
Integer index = null;
for(int j = 0;j < graph[course].size();j++){
index = (Integer)(graph[course].get(j));
if(used[index]) return false;
if(global[index]) continue;
global[index] = true;
used[index] = true;
stack.addLast(index);
flag = true;
break;
}
if(!flag) {index = stack.removeLast();used[index] = false;}
}
}
return true;
}
}2.Course Schedule II
There are a total of n courses you have to take, labeled from0 ton - 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 prerequisitepairs, return the ordering of courses you should take to finish all courses.
There may be multiple correct orders, you just need to return one of them. If it is impossible to finish all courses, return an empty array.
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 the correct course order is[0,1]
4, [[1,0],[2,0],[3,1],[3,2]]
There are a total of 4 courses to take. To take course 3 you should have finished both courses 1 and 2. Both courses 1 and 2 should be taken after you finished course 0. So one correct course order is[0,1,2,3]. Another
correct ordering is[0,2,1,3].
Note:
The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more abouthow
a graph is represented.
程式碼如下：
public class Solution {
public int[] findOrder(int numCourses, int[][] prerequisites) {
ArrayList[] g = new ArrayList[numCourses];
for(int[] pre:prerequisites){
if(g[pre[0]] == null) g[pre[0]] = new ArrayList<Integer>();
g[pre[0]].add(pre[1]);
}
int k = 0;
int cur = 0;
int[] res = new int[numCourses];
if(prerequisites.length == 0 && numCourses > 0){
for(int i = 0;i < numCourses;i++) res[i] = i;
return res;
}
LinkedList<Integer> stack = new LinkedList<Integer>();
boolean[] global = new boolean[numCourses];
for(int i = 0;i < numCourses;i++){
if(global[i]) continue;
global[i] = true;
if(g[i] == null) {res[k++] = i; continue;}
boolean[] used = new boolean[numCourses];
stack.addLast(i);
used[i] = true;
while(stack.size() > 0){
int cour = stack.getLast();
if(g[cour] == null){cur = stack.removeLast(); used[cur] = false; res[k++] = cur; continue;}
boolean flag = false;
for(int index = 0;index < g[cour].size();index++){
int c = (Integer)(g[cour].get(index));
if(used[c]) return new int[0];
if(global[c]) continue;
used[c] = true;
global[c] = true;
stack.addLast(c);
flag = true;
break;
}
if(!flag){cur = stack.removeLast(); used[cur] = false; res[k++] = cur;}
}
}//
return res;
}
}