如果我有[1,2,3,4]和k = 3这样的数组，那么输出应该是 [1,2,3] [2,3,4] [1,3,4]按排序顺序。
我可以在k = 2时这样做，但是不能想到一种方法来为k的任何值更通用。
final int arr[] = new int[] { 1, 2, 3, 4 };
final int max = 2;
for (int i = 0; i < arr.length - 1; i++) {
for (int j = i + 1; j < arr.length; j++) {
for (int k = 0; k < max; k = k + 2) {
System.out.println(i + "" + j);
}
}
}
答案 0 :(得分：0)
正如我上面所说，尝试将数组转换为ArrayList，对其进行排序。 想象一下，现在你有一张包含整数值列表的原始论文。 您复制此文件，交叉复制中的一个整数值并将此文件放入您的文件夹（以下代码中的HashSet）。然后再复制原件，依此类推。
package Subarrays;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashSet;
import java.util.List;
import java.util.Scanner;
import java.util.Set;
public class Subarrays {
public static void main(String[] arg) {
int arraySize = 0;
int initElementNumber = 0;
int initElementValue = 0;
Scanner scanner = new Scanner(System.in);
// Creates an array
System.out.println("Type the size of the array: ");
do {
try {
arraySize = scanner.nextInt();
} catch (Exception e) {
e.printStackTrace();
}
} while (arraySize == 0);
System.out.println("Type the number of element, which you want to initialize yourself: ");
do {
try {
initElementNumber = scanner.nextInt();
} catch (Exception e) {
e.printStackTrace();
}
} while (initElementNumber == 0);
System.out.println("Type value of the element: ");
do {
try {
initElementValue = scanner.nextInt();
} catch (Exception e) {
e.printStackTrace();
}
} while (initElementValue == 0);
Integer arr[] = new Integer[arraySize];
arr[initElementNumber] = initElementValue;
for (int i = 0; i < arr.length; i++) {
if (i != initElementNumber) {
arr[i] = i;
}
}
// Converts the array to arrayList
ArrayList<Integer> list = new ArrayList<Integer>(Arrays.asList(arr));
// Sorts the list
Collections.sort(list);
// Creates a clone of the list
ArrayList<Integer> listTemp = (ArrayList<Integer>) list.clone();
// Creates a set of the future lists
Set<ArrayList<Integer>> set = new HashSet<ArrayList<Integer>>();
// Iterates over the list and removes one element
for (int i = 0; i < list.size(); i++) {
listTemp.remove(i); // listTemp restructured here
// Adding listItem to set
set.add(listTemp);
// Creates new clone of the list
listTemp = (ArrayList<Integer>) list.clone();
}
// Writes content of the set to the console
for (List<Integer> list2 : set) {
System.out.print("List: ");
for (Integer integer2 : list2) {
System.out.print(" " + integer2 + " ");
}
System.out.println();
}
}
}
执行代码的结果将是：
Type the size of the array:
5
Type the number of element, which you want to initialize yourself:
2
Type value of the element:
8
List: 0 3 4 8
List: 0 1 4 8
List: 1 3 4 8
List: 0 1 3 8
List: 0 1 3 4
<强>更新 我更仔细地阅读了你的问题，并决定重写满足主要要求的代码：在大小为n的数组中查找大小为k的所有排序子数组。
package Subarrays;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashSet;
import java.util.List;
import java.util.Scanner;
import java.util.Set;
public class Subarrays2 {
private static int ARRAY_SIZE = 0;
private static int SIZE_OF_SUBARRAYS = 0;
private static Set<ArrayList<Integer>> set = new HashSet<ArrayList<Integer>>();
public static void main(String[] arg) {
Scanner scanner = new Scanner(System.in);
// Creates an array
System.out.println("Type the size of the array: ");
do {
try {
ARRAY_SIZE = scanner.nextInt();
} catch (Exception e) {
e.printStackTrace();
}
} while (ARRAY_SIZE == 0);
System.out.println("Type the size of subarrays: ");
do {
try {
SIZE_OF_SUBARRAYS = scanner.nextInt();
} catch (Exception e) {
e.printStackTrace();
}
} while (SIZE_OF_SUBARRAYS == 0);
scanner.close();
Integer arr[] = new Integer[ARRAY_SIZE];
for (int i = 0; i < ARRAY_SIZE; i++) {
arr[i] = i;
}
// Converts the array to arrayList
ArrayList<Integer> list = new ArrayList<Integer>(Arrays.asList(arr));
// Sorts the list
Collections.sort(list);
recursion(list); // This method work until find all of the sorted lists
// with size SIZE_OF_SUBARRAYS in array with size ARRAY_SIZE
// Writes content of the set to the console
for (List<Integer> list2 : set) {
System.out.print("List: ");
for (Integer integer2 : list2) {
System.out.print(" " + integer2 + " ");
}
System.out.println();
}
}
private static void recursion(ArrayList<Integer> notSmallestCombination) {
ArrayList<Integer> listTemp = (ArrayList<Integer>) notSmallestCombination.clone();
for (int i = 0; i < notSmallestCombination.size(); i++) {
listTemp.remove(i);
if (listTemp.size() == SIZE_OF_SUBARRAYS) {
for (int j = 0; j < notSmallestCombination.size(); j++) {
ArrayList<Integer> list2Temp = (ArrayList<Integer>) notSmallestCombination.clone();
list2Temp.remove(j);
set.add(list2Temp);
}
} else {
recursion(listTemp);
}
listTemp = (ArrayList<Integer>) notSmallestCombination.clone();
}
}
}
结果是：
Type the size of the array:
5
Type the size of subarrays:
3
List: 0 3 4
List: 0 2 3
List: 0 1 2
List: 1 3 4
List: 1 2 3
List: 0 2 4
List: 0 1 3
List: 2 3 4
List: 1 2 4
List: 0 1 4
答案 1 :(得分：-1)
代码使用置换函数：
// Java program to print all combination of size r in an array of size n
import java.io.*;
class Permutation {
/* 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(int arr[], int data[], int start,
int end, int index, int r)
{
// 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("");
return;
}
// replace index with all possible elements. The condition
// "end-i+1 >= r-index" makes sure that including one element
// at index will make a combination with remaining elements
// at remaining positions
for (int i=start; i<=end && end-i+1 >= r-index; i++)
{
data[index] = arr[i];
combinationUtil(arr, data, i+1, end, index+1, r);
}
}
// The main function that prints all combinations of size r
// in arr[] of size n. This function mainly uses combinationUtil()
static void printCombination(int arr[], int n, int r)
{
// A temporary array to store all combination one by one
int data[]=new int[r];
// Print all combination using temprary array 'data[]'
combinationUtil(arr, data, 0, n-1, 0, r);
}
/*Driver function to check for above function*/
public static void main (String[] args) {
int arr[] = {1, 2, 3, 4, 5};
int r = 3;
int n = arr.length;
printCombination(arr, n, r);
}
}
/* This code is contributed by Devesh Agrawal */
一旦你完成所有的排列，你只需要用它构建一个数组