我需要随机洗牌以下数组:
int[] solutionArray = {1, 2, 3, 4, 5, 6, 6, 5, 4, 3, 2, 1};
有什么函数可以做到吗?
当前回答
下面是使用集合的完整解决方案。洗牌的方法:
public static void shuffleArray(int[] array) {
List<Integer> list = new ArrayList<>();
for (int i : array) {
list.add(i);
}
Collections.shuffle(list);
for (int i = 0; i < list.size(); i++) {
array[i] = list.get(i);
}
}
请注意，由于Java无法在int[]和Integer[]之间平滑转换(因此int[]和List<Integer>)，它受到了影响。
其他回答
你有几个选择。当涉及到洗牌时，列表与数组有点不同。
如下所示，数组比列表快，而基元数组比对象数组快。
样品时间
List<Integer> Shuffle: 43133ns
Integer[] Shuffle: 31884ns
int[] Shuffle: 25377ns
下面是shuffle的三种不同实现。您应该只使用集合。如果处理的是集合，则Shuffle。没有必要为了排序而将数组包装到一个集合中。下面的方法实现起来非常简单。
ShuffleUtil Class
import java.lang.reflect.Array;
import java.util.*;
public class ShuffleUtil<T> {
private static final int[] EMPTY_INT_ARRAY = new int[0];
private static final int SHUFFLE_THRESHOLD = 5;
private static Random rand;
主要方法
public static void main(String[] args) {
List<Integer> list = null;
Integer[] arr = null;
int[] iarr = null;
long start = 0;
int cycles = 1000;
int n = 1000;
// Shuffle List<Integer>
start = System.nanoTime();
list = range(n);
for (int i = 0; i < cycles; i++) {
ShuffleUtil.shuffle(list);
}
System.out.printf("%22s: %dns%n", "List<Integer> Shuffle", (System.nanoTime() - start) / cycles);
// Shuffle Integer[]
start = System.nanoTime();
arr = toArray(list);
for (int i = 0; i < cycles; i++) {
ShuffleUtil.shuffle(arr);
}
System.out.printf("%22s: %dns%n", "Integer[] Shuffle", (System.nanoTime() - start) / cycles);
// Shuffle int[]
start = System.nanoTime();
iarr = toPrimitive(arr);
for (int i = 0; i < cycles; i++) {
ShuffleUtil.shuffle(iarr);
}
System.out.printf("%22s: %dns%n", "int[] Shuffle", (System.nanoTime() - start) / cycles);
}
变换通用列表
// ================================================================
// Shuffle List<T> (java.lang.Collections)
// ================================================================
@SuppressWarnings("unchecked")
public static <T> void shuffle(List<T> list) {
if (rand == null) {
rand = new Random();
}
int size = list.size();
if (size < SHUFFLE_THRESHOLD || list instanceof RandomAccess) {
for (int i = size; i > 1; i--) {
swap(list, i - 1, rand.nextInt(i));
}
} else {
Object arr[] = list.toArray();
for (int i = size; i > 1; i--) {
swap(arr, i - 1, rand.nextInt(i));
}
ListIterator<T> it = list.listIterator();
int i = 0;
while (it.hasNext()) {
it.next();
it.set((T) arr[i++]);
}
}
}
public static <T> void swap(List<T> list, int i, int j) {
final List<T> l = list;
l.set(i, l.set(j, l.get(i)));
}
public static <T> List<T> shuffled(List<T> list) {
List<T> copy = copyList(list);
shuffle(copy);
return copy;
}
变换泛型数组
// ================================================================
// Shuffle T[]
// ================================================================
public static <T> void shuffle(T[] arr) {
if (rand == null) {
rand = new Random();
}
for (int i = arr.length - 1; i > 0; i--) {
swap(arr, i, rand.nextInt(i + 1));
}
}
public static <T> void swap(T[] arr, int i, int j) {
T tmp = arr[i];
arr[i] = arr[j];
arr[j] = tmp;
}
public static <T> T[] shuffled(T[] arr) {
T[] copy = Arrays.copyOf(arr, arr.length);
shuffle(copy);
return copy;
}
打乱原语数组
// ================================================================
// Shuffle int[]
// ================================================================
public static <T> void shuffle(int[] arr) {
if (rand == null) {
rand = new Random();
}
for (int i = arr.length - 1; i > 0; i--) {
swap(arr, i, rand.nextInt(i + 1));
}
}
public static <T> void swap(int[] arr, int i, int j) {
int tmp = arr[i];
arr[i] = arr[j];
arr[j] = tmp;
}
public static int[] shuffled(int[] arr) {
int[] copy = Arrays.copyOf(arr, arr.length);
shuffle(copy);
return copy;
}
实用方法
简单的实用工具方法复制和转换数组到列表，反之亦然。
// ================================================================
// Utility methods
// ================================================================
protected static <T> List<T> copyList(List<T> list) {
List<T> copy = new ArrayList<T>(list.size());
for (T item : list) {
copy.add(item);
}
return copy;
}
protected static int[] toPrimitive(Integer[] array) {
if (array == null) {
return null;
} else if (array.length == 0) {
return EMPTY_INT_ARRAY;
}
final int[] result = new int[array.length];
for (int i = 0; i < array.length; i++) {
result[i] = array[i].intValue();
}
return result;
}
protected static Integer[] toArray(List<Integer> list) {
return toArray(list, Integer.class);
}
protected static <T> T[] toArray(List<T> list, Class<T> clazz) {
@SuppressWarnings("unchecked")
final T[] arr = list.toArray((T[]) Array.newInstance(clazz, list.size()));
return arr;
}
范围类
生成一个值范围，类似于Python的range函数。
// ================================================================
// Range class for generating a range of values.
// ================================================================
protected static List<Integer> range(int n) {
return toList(new Range(n), new ArrayList<Integer>());
}
protected static <T> List<T> toList(Iterable<T> iterable) {
return toList(iterable, new ArrayList<T>());
}
protected static <T> List<T> toList(Iterable<T> iterable, List<T> destination) {
addAll(destination, iterable.iterator());
return destination;
}
protected static <T> void addAll(Collection<T> collection, Iterator<T> iterator) {
while (iterator.hasNext()) {
collection.add(iterator.next());
}
}
private static class Range implements Iterable<Integer> {
private int start;
private int stop;
private int step;
private Range(int n) {
this(0, n, 1);
}
private Range(int start, int stop) {
this(start, stop, 1);
}
private Range(int start, int stop, int step) {
this.start = start;
this.stop = stop;
this.step = step;
}
@Override
public Iterator<Integer> iterator() {
final int min = start;
final int max = stop / step;
return new Iterator<Integer>() {
private int current = min;
@Override
public boolean hasNext() {
return current < max;
}
@Override
public Integer next() {
if (hasNext()) {
return current++ * step;
} else {
throw new NoSuchElementException("Range reached the end");
}
}
@Override
public void remove() {
throw new UnsupportedOperationException("Can't remove values from a Range");
}
};
}
}
}
无随机解:
static void randomArrTimest(int[] some){
long startTime = System.currentTimeMillis();
for (int i = 0; i < some.length; i++) {
long indexToSwap = startTime%(i+1);
long tmp = some[(int) indexToSwap];
some[(int) indexToSwap] = some[i];
some[i] = (int) tmp;
}
System.out.println(Arrays.toString(some));
}
使用ArrayList<Integer>可以帮助您解决洗牌问题，而无需应用大量逻辑和消耗更少的时间。以下是我的建议:
ArrayList<Integer> x = new ArrayList<Integer>();
for(int i=1; i<=add.length(); i++)
{
x.add(i);
}
Collections.shuffle(x);
下面是一个使用Apache Commons Math 3的解决方案。X(仅适用于int[]数组):
MathArrays.shuffle(array);
http://commons.apache.org/proper/commons-math/javadocs/api-3.6.1/org/apache/commons/math3/util/MathArrays.html shuffle (int [])
另外，Apache Commons Lang 3.6为ArrayUtils类引入了新的shuffle方法(用于对象和任何基本类型)。
ArrayUtils.shuffle(array);
http://commons.apache.org/proper/commons-lang/javadocs/api-release/org/apache/commons/lang3/ArrayUtils.html#shuffle-int:A-
Box from int[] to List<Integer> 用集合洗牌。洗牌法
int[] solutionArray = { 1, 2, 3, 4, 5, 6, 6, 5, 4, 3, 2, 1 };
List<Integer> list = Arrays.stream(solutionArray).boxed().collect(Collectors.toList());
Collections.shuffle(list);
System.out.println(list.toString());
// [1, 5, 5, 4, 2, 6, 1, 3, 3, 4, 2, 6]
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