的元素的最近我实现的代码为下面 假设我有2个数组作为 -组合多个阵列
arr1 = [a,b,c]
arr2 = [d,e]
和应该尽可能
output = [ad,ae,bd,be,cd,ce]
现在假设我有多个阵列。
为如：
arr1=[a,b,c]
arr2=[d,e]
arr3=[f,g,h,i]
arrN=[x,y,z,h,o]
output = [adf..x,adf..y and so on]
如何实现这一点JAVA？好心帮
我尝试：
for(int i=0;i<arr1.length();i++)
{
for(int j=0;i<arr2.length();j++)
{
System.out.print(arr1[i] + arr2[j]);
}
}
这是我在来自多个阵列（容器）得到的组合的尝试。
private List<List<T>> getCombination(int currentIndex, List<TempContainer<T>> containers) {
if (currentIndex == containers.size()) {
// Skip the items for the last container
List<List<T>> combinations = new ArrayList<List<T>>();
combinations.add(new ArrayList<T>());
return combinations;
}
List<List<T>> combinations = new ArrayList<List<T>>();
TempContainer<T> container = containers.get(currentIndex);
List<T> containerItemList = container.getItems();
// Get combination from next index
List<List<T>> suffixList = getCombination(currentIndex + 1, containers);
int size = containerItemList.size();
for (int ii = 0; ii < size; ii++) {
T containerItem = containerItemList.get(ii);
if (suffixList != null) {
for (List<T> suffix : suffixList) {
List<T> nextCombination = new ArrayList<T>();
nextCombination.add(containerItem);
nextCombination.addAll(suffix);
combinations.add(nextCombination);
}
}
}
return combinations;
}
该方法采用包含项目的容器列表。例如容器1将具有[a，b]，容器2将具有[c，d，e]。该方法可以采用任意数量的容器。 //容器类被声明如下：
public class TempContainer<T> {
private List<T> items;
public void setItems(List<T> items) {
this.items = items;
}
public List<T> getItems() {
return items;
}
}
调用此方法如下：不同类型的
List<String> list1 = new ArrayList<String>();
list1.add("1");
list1.add("2");
List<String> list2 = new ArrayList<String>();
list2.add("3");
list2.add("4");
list2.add("5");
TempContainer container1 = new TempContainer();
container1.setItems(list1);
TempContainer container2 = new TempContainer();
container2.setItems(list2);
List<TempContainer> containers = new ArrayList<TempContainer>(2);
containers.add(container1);
containers.add(container2);
// Get combinations
List<List<T>> combinations = getCombination(0, containers);
它不适用于4个容器。 –
@nazar_art你用4个容器得到什么问题？我已经尝试过4个和更多的容器，它已经工作。 – Priyesh
它因'NullPointerException'失败。 –
只是简单 -
ArrayList<Integer> combination(int[] ar1, int[] ar2)
{
ArrayList<Integer> result = new ArrayList<int>();
for (int i1 : ar1)
{
for (int i2 : ar2)
{
result.add(i1 * i2);
}
}
return result;
}
嗯，我认为递归是解决一般情况。
import java.util.*;
class Ideone
{
public static void main (String[] args) throws java.lang.Exception
{
ArrayList<int[]> arrays = new ArrayList<int[]>();
arrays.add(new int[] { 1, 2, 3 });
arrays.add(new int[] { 4, 5 });
arrays.add(new int[] { 6, 7 });
ArrayList<Integer> result = combination(arrays);
for (int i : result)
{
System.out.println(i);
}
}
static ArrayList<Integer> combination(ArrayList<int[]> arrays)
{
if (arrays.size() == 0)
return null;
ArrayList<Integer> result = new ArrayList<Integer>();
combination_sub(arrays, 0, result);
return result;
}
static void combination_sub(ArrayList<int[]> arrays, int idx, ArrayList<Integer> result)
{
if (arrays.size() == idx)
{
result.add(1);
return;
}
int[] ar = arrays.get(idx);
ArrayList<Integer> sub_result = new ArrayList<Integer>();
combination_sub(arrays, idx + 1, sub_result);
for (int i : ar)
{
for (int sub_i : sub_result)
{
result.add(i * sub_i);
}
}
}
}
什么具有成阵列的阵列：
String[] arr1 = { "a", "b", "c" };
String[] arr2 = { "d", "e" };
String[] arr3 = { "f", "g", "h", "i" };
String[][] arrays = new String[][] { arr1, arr2, arr3 };
public void mix(String[][] arrays) {
if (arrays.length > 0) {
List<String> result = Arrays.asList(arrays[0]);
for (int i = 1; i < arrays.length - 1; i++) {
String[] aux = arrays[i];
for (int j = 0; j < aux.length; j++) {
result.add(j, result.get(j) + aux[j]);
}
}
}
}
我有类似的要求，但对集合。这是Priyesh解决方案的修改版本。
注意：我删除了依赖于TempContainer<T>类。
public class TestCombinations {
@Test
public void test() {
final List<Object> l1 = new ArrayList<Object>();
l1.add(1);
l1.add(2);
final List<Object> l2 = new ArrayList<Object>();
l2.add("3");
l2.add("4");
l2.add("5");
final List<Object> l3 = new ArrayList<Object>();
l3.add(true);
l3.add(false);
final List<List<Object>> c = new ArrayList<>(3);
c.add(l1);
c.add(l2);
c.add(l3);
// Get combinations
final List<List<Object>> m = this.combine(c);
System.out.println(m);
}
public List<List<Object>> combine(final List<List<Object>> containers) {
return this.combineInternal(0, containers);
}
private List<List<Object>> combineInternal(final int currentIndex,
final List<List<Object>> containers) {
if (currentIndex == containers.size()) {
// Skip the items for the last container
final List<List<Object>> combinations = new ArrayList<>();
combinations.add(Collections.emptyList());
return combinations;
}
final List<List<Object>> combinations = new ArrayList<>();
final List<Object> containerItemList = containers.get(currentIndex);
// Get combination from next index
final List<List<Object>> suffixList = this.combineInternal(
currentIndex + 1, containers);
final int size = containerItemList.size();
for (int i = 0; i < size; i++) {
final Object containerItem = containerItemList.get(i);
if (suffixList != null) {
for (final List<Object> suffix : suffixList) {
final List<Object> nextCombination = new ArrayList<>();
nextCombination.add(containerItem);
nextCombination.addAll(suffix);
combinations.add(nextCombination);
}
}
}
return combinations;
}
}
发布您的尝试。 – Maroun
使用4个循环有什么问题？ –
请参阅编辑 – ikh