例如，我有一个包含一些讲座实例的列表，每个讲座有一定数量的学生参加本次讲座，还有一个包含一些课堂实例的列表，每个教室都有最大的容量。
现在，我打算为演讲列表中的每个演讲分配一个教室列表中的教室，演讲课中的所有演讲都应该有一个教室，然后创建一个地图来存储这种可能性。
我想以集合的形式返回所有这些可能的匹配。
例如:
Classroom List: [Classroom1(50),Classroom2(70),Classroom3(80)]
Lecture list: [Lecture1(50), Lecture2(70), Lecture3(50)]
然后，我们有3种可能的地图，分别是:
{lecture1:classroom1, lecture2:classroom2, lecture3:classroom3} and
{lecture1:classroom1, lecture2:classroom3, lecture3:classroom2} and
{lecture1:classroom2, lecture2:classroom3, lecture3:classroom1}
之后，应将所有可能的地图存储在一个集中。
我是编程新手，还没有学习算法，也许这就是为什么我在这个问题上如此努力，如果有人可以帮助我解决这个问题，我将不胜感激。
最佳答案
所以我深陷于此，并写了一个可行的解决方案
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
class ClassMatcher {
//The set of all possible matchings.
static ArrayList<ArrayList<Pair>> set = new ArrayList<ArrayList<Pair>>();
// The current matching being built
static ArrayList<Pair> cur = new ArrayList<Pair>();
public static void main(String[] args) {
Lecture[] l = { new Lecture(50, 1), new Lecture(70, 2), new Lecture(50, 3)};
ArrayList<Classroom> c = new ArrayList<>(Arrays.asList(
new Classroom(50, 1), new Classroom(70, 2),
new Classroom(100, 3)));
for (int i = 0; i < l.length; i++) {
//Fill with dummy values
cur.add(new Pair(new Classroom(-1, -1), new Lecture(-1, -1)));
}
// Sort the arrays to save work in rec()
Arrays.sort(l);
//Sort classrooms in descending order
Collections.sort(c, new Comparator<Classroom>() {
@Override
public int compare(Classroom o1, Classroom o2) {
return o1.compareTo(o2) * -1;
}
});
recursive(l, c, 0);
// Print all the sets
for (int i = 0; i < set.size(); i++) {
System.out.print("{");
for (int j = 0; j < set.get(i).size(); j++) {
System.out.print("Lecture " + set.get(i).get(j).l + ": "
+ "Classroom " + set.get(i).get(j).c);
if (j < set.get(i).size() - 1) {
System.out.print(", ");
} else {
System.out.print("}");
}
}
System.out.println();
}
}
public static void recursive(Lecture[] lectureList,
ArrayList<Classroom> classroomList, int curLecture) {
for (int i = 0; i < classroomList.size(); i++) {
// if the classroom is smaller than the lecture we cna stop as the
// lists are sorted so all other lectures will be to big for the
// current classroom
if (lectureList[curLecture].size > classroomList.get(i).size) {
return;
}
//Match the current classroom to the current lecture and add to the working matching
cur.set(curLecture, new Pair(classroomList.get(i), lectureList[curLecture]));
//If there are more lectures to do then remove the used classroom and recursively call.
if (curLecture < lectureList.length - 1) {
Classroom tmp = classroomList.remove(i);
recursive(lectureList, classroomList, curLecture + 1);
classroomList.add(i, tmp);
}
// If no Lectures left then add this matching to the set of all matchings.
else {
ArrayList<Pair> copy = (ArrayList<Pair>) cur.clone();
set.add(copy);
}
}
}
}
class Classroom implements Comparable<Classroom> {
int size;
int number;
public Classroom(int s, int n) {
size = s;
number = n;
}
@Override
public int compareTo(Classroom o) {
return Integer.compare(this.size, o.size);
}
public String toString() {
return number + " (" + size + ")";
}
}
class Lecture implements Comparable<Lecture> {
int size;
int number;
public Lecture(int s, int n) {
size = s;
number = n;
}
@Override
public int compareTo(Lecture o) {
return Integer.compare(this.size, o.size);
}
public String toString() {
return number + " (" + size + ")";
}
}
class Pair {
Classroom c;
Lecture l;
public Pair(Classroom c, Lecture l) {
this.c = c;
this.l = l;
}
}
这给出了输出
{Lecture 1 (50): Classroom 3 (100), Lecture 3 (50): Classroom 1 (50), Lecture 2 (70): Classroom 2 (70)}
{Lecture 1 (50): Classroom 2 (70), Lecture 3 (50): Classroom 1 (50), Lecture 2 (70): Classroom 3 (100)}
{Lecture 1 (50): Classroom 1 (50), Lecture 3 (50): Classroom 3 (100), Lecture 2 (70): Classroom 2 (70)}
{Lecture 1 (50): Classroom 1 (50), Lecture 3 (50): Classroom 2 (70), Lecture 2 (70): Classroom 3 (100)}