Java-从多个数组中获取n个元素的组合
我有4种类型的整数值。我想从下面的数组中生成由3个元素组成的所有可能的组合,如Java-从多个数组中获取n个元素的组合,java,collections,combinations,Java,Collections,Combinations,我有4种类型的整数值。我想从下面的数组中生成由3个元素组成的所有可能的组合,如 5 1 72 3 7 9 8 14 11 //etc List<Integer> list1 = Arrays.asList(5, 7, 11, 2, 10); List<Integer> list2 = Arrays.asList(1, 9, 25); List<Integer> list3 = Arrays.asList(72, 8); List<Integer>
5 1 72
3 7 9
8 14 11 //etc
List<Integer> list1 = Arrays.asList(5, 7, 11, 2, 10);
List<Integer> list2 = Arrays.asList(1, 9, 25);
List<Integer> list3 = Arrays.asList(72, 8);
List<Integer> list4 = Arrays.asList(3, 14, 22, 37, 19);
如何获得3个元素的组合
private static List<List<Integer>> getCombination(int currentIndex, List<TempContainer<Integer>> containers) {
if (currentIndex == containers.size()) {
// Skip the items for the last container
List<List<Integer>> combinations = new ArrayList<>();
combinations.add(new ArrayList<>());
return combinations;
}
List<List<Integer>> combinations = new ArrayList<>();
TempContainer<Integer> container = containers.get(currentIndex);
List<Integer> containerItemList = container.getItems();
List<List<Integer>> suffixList = getCombination(currentIndex + 1, containers);
int size = containerItemList.size();
for (int ii = 0; ii < size; ii++) {
Integer containerItem = containerItemList.get(ii);
if (suffixList != null) {
for (List<Integer> suffix : suffixList) {
List<Integer> nextCombination = new ArrayList<>();
nextCombination.add(containerItem);
nextCombination.addAll(suffix);
combinations.add(nextCombination);
}
}
}
return combinations;
}
TempContainer container1 = new TempContainer();
container1.setItems(list1);
TempContainer container2 = new TempContainer();
container2.setItems(list2);
TempContainer container3 = new TempContainer();
container3.setItems(list3);
TempContainer container4 = new TempContainer();
container4.setItems(list4);
List<TempContainer<Integer>> containers = new ArrayList<>(3);
containers.add(container1);
containers.add(container2);
containers.add(container3);
containers.add(container4);
// Get combinations
List<List<Integer>> combinations = getCombination(0, containers);
私有静态列表getCombination(int-currentIndex,列表容器){
if(currentIndex==containers.size()){
//跳过最后一个容器的项目
列表组合=新的ArrayList();
添加(新的ArrayList());
收益组合;
}
列表组合=新的ArrayList();
TempContainer=containers.get(currentIndex);
List containerItemList=container.getItems();
列表后缀列表=getCombination(currentIndex+1,容器);
int size=containerItemList.size();
用于(int ii=0;ii
基本上,你必须做你以前做过的事情。唯一的区别是,由于您现在只使用四个列表中的三个,因此必须添加第二个置换循环,循环遍历相关列表的所有可能组合。因此,算法是:
- 从四个列表中生成三个列表的所有可能组合
- 对于每个组合,使用前面的算法(用于生成所有可能的元素组合)
- 从四个列表中生成三个列表的所有可能组合
- 对于每个组合,使用前面的算法(用于生成所有可能的元素组合)
- 我认为更好的解决方案是:
public static void main(String[] args) {
List<Integer> list1 = Arrays.asList(5, 7, 11, 2, 10);
List<Integer> list2 = Arrays.asList(1, 9, 25);
List<Integer> list3 = Arrays.asList(72, 8);
List<Integer> list4 = Arrays.asList(3, 14, 22, 37, 19);
List<Integer[]> combinations = getCombinations(list1, list2, list3, list4);
for (Integer[] combination : combinations)
System.out.println(combination[0] + "," + combination[1] + "," + combination[2]);
}
public static List<Integer[]> getCombinations(List<Integer>... lists) {
return _getCombination(0, new ArrayList<>(), new ArrayList<>(), lists);
}
private static List<Integer[]> _getCombination(int depth, List<Integer[]> currentCombinations,
List<Integer> currentCombination, List<Integer>... lists) {
if (currentCombination.size() == 3) {
currentCombinations.add(currentCombination.toArray(new Integer[3]));
} else {
for (int i = 0 ; i < lists.length ; i++)
for (int j = 0 ; j < lists[i].size() ; j++) {
currentCombination.add(lists[i].get(j));
_getCombination(depth + 1, currentCombinations, currentCombination, lists);
currentCombination.remove(lists[i].get(j));
}
}
return currentCombinations;
}
publicstaticvoidmain(字符串[]args){
list1=Arrays.asList(5,7,11,2,10);
list2=Arrays.asList(1,9,25);
list3=Arrays.asList(72,8);
list4=Arrays.asList(3,14,22,37,19);
列表组合=获取组合(列表1、列表2、列表3、列表4);
对于(整数[]组合:组合)
System.out.println(组合[0]+“,“+组合[1]+”,“+组合[2]);
}
公共静态列表组合(列表…列表){
返回_getCombination(0,new ArrayList(),new ArrayList(),List);
}
私有静态列表_getCombination(整数深度,列表当前组合,
列表当前组合,列表…列表){
if(currentCombination.size()==3){
currentcombings.add(currentCombination.toArray(新整数[3]);
}否则{
for(int i=0;i
我认为更好的解决方案是:
public static void main(String[] args) {
List<Integer> list1 = Arrays.asList(5, 7, 11, 2, 10);
List<Integer> list2 = Arrays.asList(1, 9, 25);
List<Integer> list3 = Arrays.asList(72, 8);
List<Integer> list4 = Arrays.asList(3, 14, 22, 37, 19);
List<Integer[]> combinations = getCombinations(list1, list2, list3, list4);
for (Integer[] combination : combinations)
System.out.println(combination[0] + "," + combination[1] + "," + combination[2]);
}
public static List<Integer[]> getCombinations(List<Integer>... lists) {
return _getCombination(0, new ArrayList<>(), new ArrayList<>(), lists);
}
private static List<Integer[]> _getCombination(int depth, List<Integer[]> currentCombinations,
List<Integer> currentCombination, List<Integer>... lists) {
if (currentCombination.size() == 3) {
currentCombinations.add(currentCombination.toArray(new Integer[3]));
} else {
for (int i = 0 ; i < lists.length ; i++)
for (int j = 0 ; j < lists[i].size() ; j++) {
currentCombination.add(lists[i].get(j));
_getCombination(depth + 1, currentCombinations, currentCombination, lists);
currentCombination.remove(lists[i].get(j));
}
}
return currentCombinations;
}
publicstaticvoidmain(字符串[]args){
list1=Arrays.asList(5,7,11,2,10);
list2=Arrays.asList(1,9,25);
list3=Arrays.asList(72,8);
list4=Arrays.asList(3,14,22,37,19);
列表组合=获取组合(列表1、列表2、列表3、列表4);
对于(整数[]组合:组合)
System.out.println(组合[0]+“,“+组合[1]+”,“+组合[2]);
}
公共静态列表组合(列表…列表){
返回_getCombination(0,new ArrayList(),new ArrayList(),List);
}
私有静态列表_getCombination(整数深度,列表当前组合,
列表当前组合,列表…列表){
if(currentCombination.size()==3){
currentcombings.add(currentCombination.toArray(新整数[3]);
}否则{
for(int i=0;i
尝试类似的方法
List<Integer> list1 = Arrays.asList(72, 8);
List<Integer> list2 = Arrays.asList(1, 9, 25);
List<Integer> list3 = Arrays.asList(5, 7, 11, 2, 10);
List<Integer> list4 = Arrays.asList(3, 14,22, 37, 19, 18);
List<List<Integer>> mainList = new ArrayList<>();
mainList.add(list4);
mainList.add(list3);
mainList.add(list2);
mainList.add(list1);
int max = 0;
// FIND MAX LIST So that We can ITERATE on that
for (int k = 0; k < mainList.size(); k++) {
if (mainList.get(k).size() > max) {
max = mainList.get(k).size();
}
}
for (int k = 0; k < max; k++) {
String string = new String(); // You can use another list of array to store combination. For testing I have used String
try {
for (int l = 0; l < mainList.size(); l++) {
if (mainList.get(l).size() > k)
string = string.concat(" " + mainList.get(l).get(k));
}
System.out.println(string);
} catch (Exception e) {
System.out.println("Error IN CALL");
}
}
list1=Arrays.asList(72,8);
list2=Arrays.asList(1,9,25);
list3=Arrays.asList(5,7,11,2,10);
list4=Arrays.asList(3,14,22,37,19,18);
List mainList=new ArrayList();
mainList.add(列表4);
mainList.add(列表3);
mainList.add(列表2);
mainList.add(列表1);
int max=0;
//查找最大列表,以便我们可以对其进行迭代
对于(int k=0;kmax){
最大值