最近我实现了以下代码 假设我有2个数组 -
arr1 = [a,b,c]
arr2 = [d,e]
应该是
output = [ad,ae,bd,be,cd,ce]
现在假设我有多个阵列。
例如：
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]
如何在JAVA中实现这一点？请帮助
我的尝试：
for(int i=0;i<arr1.length();i++)
{
for(int j=0;i<arr2.length();j++)
{
System.out.print(arr1[i] + arr2[j] );
}
}
答案 0 :(得分：3)
这是我尝试从多个数组（容器）中获取组合。
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;
}
该方法获取带有项目的容器列表。例如容器1将具有[a，b]，而容器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;
}
}
按如下方式调用此方法：
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);
答案 1 :(得分：1)
将数组放在数组中怎么样：
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]);
}
}
}
}
答案 2 :(得分：1)
我有类似的要求，但是对于不同类型的集合。这是Priyesh解决方案的修改版本。
注意：我删除了对TempContainer<T>类的依赖。
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;
}
}
答案 3 :(得分：0)
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;
}
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);
}
}
}
}
答案 4 :(得分：0)
您可以在java 8 Stream Api中轻松实现此目的：
final答案 5 :(得分：0)
如果是3个数组。如果要3个数组以外的其他数组，可以插入其他代码。递归解决方案
/* This code is contributed by Devesh Agrawal and modified for personal solution by prism */
/* contributed by Devesh Agrawal at https://www.geeksforgeeks.org/print-all-possible-combinations-of-r-elements-in-a-given-array-of-size-n/ */
package com.practice.probabilty;
import java.util.ArrayList;
public class CombinationNumber {
public static void main(String[] args) {
int[] arr1 = new int[] {1, 2, 3, 4, 5, 6, 7, 8, 9 };
int[] arr2 = new int[] {0, 1, 2, 3, 4, 5, 6, 7, 8, 9 };
int[] arr3 = new int[] {0, 2, 4, 6, 8};
int[] arr = arr1.clone();
int r = 3; //amount of digits as need
int n1 = arr1.length; //amount of all digits provided for composing the number
int n2 = arr2.length;
int n3 = arr3.length;
int[] count = new int[]{0}; //for counting total amount of all combination numbers
printCombination(arr1, arr2, arr3, arr, n1, n2, n3, r, count);
System.out.println("Total amount of number = " + count[0]);
}
/*
* arr[] ---> Input Array data[] ---> Temporary array to store current
* combination start & end ---> Staring and Ending indexes in arr[] index --->
* Current index in data[] r ---> Size of a combination to be printed
*/
static void combinationUtil_1(int arr1[], int arr2[], int arr3[], int arr[],
int n1, int n2, int n3, int r, int index,
int data[], int i1, int i2, int i3, int[] count) {
// Print index and i
// System.out.print("data index = " + index + ", arr i = " + i + ", order = " + order);
// Current combination is ready to be printed, print it
if (index == r) {
for (int j = 0; j < r; j++) {
System.out.print(" " + data[j] + " ");
}
System.out.println();
count[0]++;
return;
}
// When no more elements are there to put in data[]
if (i1 >= n1 || i2 >= n2 || i3 >= n3)
return;
// current is included, put next at next location
if(index == 0)
{
data[index] = arr1[i1];
combinationUtil_1(arr1, arr2, arr3, arr2, n1, n2, n3, r, index + 1, data, i1++, i2, i3, count);
combinationUtil_1(arr1, arr2, arr3, arr2, n1, n2, n3, r, index, data, i1++, i2, i3, count);
}else if(index == 1)
{
data[index] = arr2[i2];
combinationUtil_1(arr1, arr2, arr3, arr3, n1, n2, n3, r, index + 1, data, i1, i2++, i3, count);
combinationUtil_1(arr1, arr2, arr3, arr3, n1, n2, n3, r, index, data, i1, i2++, i3, count);
}
else
{
data[index] = arr3[i3];
combinationUtil_1(arr1, arr2, arr3, arr1, n1, n2, n3, r, index + 1, data, i1, i2, i3++, count);
combinationUtil_1(arr1, arr2, arr3, arr1, n1, n2, n3, r, index, data, i1, i2, i3++, count);
}
}
// The main function that prints all combinations of size r
// This function mainly uses combinationUtil()
static void printCombination(int arr1[], int arr2[], int arr3[], int arr[],
int n1, int n2, int n3, int r, int[] count) {
// A temporary array to store all combination one by one
int data[] = new int[r];
// Print all combination using temporary array 'data[]'
combinationUtil_1(arr1, arr2, arr3, arr, n1, n2, n3, r, 0, data, 0, 0, 0, count);
}
}
答案 6 :(得分：0)
对于几个数组，在此代码中为4个数组，递归
package com.practice.probabilty;
import java.util.ArrayList;
import java.util.HashMap;
public class Example {
public static void main(String[] args) {
int[] arr1 = new int[] {1, 2, 3, 4};
int[] arr2 = new int[] {0, 1, 2, 3, 4, 5, 6, 7, 8, 9};
int[] arr3 = new int[] {0, 1, 2, 3, 4, 5, 6, 7, 8, 9};
int[] arr4 = new int[] {1, 3, 5, 7, 9};
ArrayList<int[]> listOfArray = new ArrayList<int[]>();
listOfArray.add(arr1);
listOfArray.add(arr2);
listOfArray.add(arr3);
listOfArray.add(arr4);
int sizeOfList = listOfArray.size();
int r = sizeOfList;
HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
for(int i = 0; i < sizeOfList; i++)
{
map.put(i, 0);
}
int[] count = new int[]{0}; //for counting total amount of all combination numbers
printCombination(listOfArray, sizeOfList, r, map, count);
System.out.println("Total amount of number = " + count[0]);
}
/*
* arr[] ---> Input Array data[] ---> Temporary array to store current
* combination start & end ---> Staring and Ending indexes in arr[] index --->
* Current index in data[] r ---> Size of a combination to be printed
*/
static void combinationUtil_1(ArrayList<int[]> listOfArray,
int sizeOfList, int r, int index,
int data[], HashMap<Integer, Integer> map, int[] count) {
if (index == r) {
//System.out.print("--> ");
for (int j = 0; j < r; j++) {
System.out.print(" " + data[j] + " ");
}
System.out.println();
count[0]++;
return;
}
for(int i = 0; i < sizeOfList; i++)
{
if(map.get(i) >= listOfArray.get(i).length)
return;
}
data[index] = listOfArray.get(index)[map.get(index)];
combinationUtil_1(listOfArray, sizeOfList, r, index + 1, data, map, count);
HashMap<Integer, Integer> map1 = new HashMap<Integer, Integer>();
for(int i = 0; i < sizeOfList; i++)
{
if(i == index)
{
map1.put(i, map.get(i) + 1);
}
else
{
map1.put(i, map.get(i));
}
}
combinationUtil_1(listOfArray, sizeOfList, r, index, data, map1, count);
HashMap<Integer, Integer> map2 = new HashMap<Integer, Integer>();
for(int i = 0; i < sizeOfList; i++)
{
if(i == index)
{
map2.put(i, map.get(i) + 1);
}
else
{
map2.put(i, map.get(i));
}
}
}
static void printCombination(ArrayList<int[]> listOfArray,
int sizeOfList, int r, HashMap<Integer, Integer> map, int[] count) {
// A temporary array to store all combination one by one
int data[] = new int[r];
// Print all combination using temporary array 'data[]'
combinationUtil_1(listOfArray, sizeOfList, r, 0, data, map, count);
//combinationUtil_1(arr1, arr2, arr3, arr, n1, n2, n3, r, 0, data, 0, 0, 0, count);
}
/* Driver function to check for above function */
}
/* This code is modified from Devesh Agrawal's */