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?
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
我在权衡这个非常流行的问题,因为没有人写过一个shuffle-copy版本。样式大量借鉴了Arrays.java,因为现在谁没有掠夺Java技术呢?包括泛型和int实现。
/**
* Shuffles elements from {@code original} into a newly created array.
*
* @param original the original array
* @return the new, shuffled array
* @throws NullPointerException if {@code original == null}
*/
@SuppressWarnings("unchecked")
public static <T> T[] shuffledCopy(T[] original) {
int originalLength = original.length; // For exception priority compatibility.
Random random = new Random();
T[] result = (T[]) Array.newInstance(original.getClass().getComponentType(), originalLength);
for (int i = 0; i < originalLength; i++) {
int j = random.nextInt(i+1);
result[i] = result[j];
result[j] = original[i];
}
return result;
}
/**
* Shuffles elements from {@code original} into a newly created array.
*
* @param original the original array
* @return the new, shuffled array
* @throws NullPointerException if {@code original == null}
*/
public static int[] shuffledCopy(int[] original) {
int originalLength = original.length;
Random random = new Random();
int[] result = new int[originalLength];
for (int i = 0; i < originalLength; i++) {
int j = random.nextInt(i+1);
result[i] = result[j];
result[j] = original[i];
}
return result;
}
其他回答
看看Collections类,特别是shuffle(…)。
使用集合来洗牌一个基本类型数组有点过分……
你自己实现这个函数很简单,比如使用Fisher-Yates shuffle:
import java.util.*;
import java.util.concurrent.ThreadLocalRandom;
class Test
{
public static void main(String args[])
{
int[] solutionArray = { 1, 2, 3, 4, 5, 6, 16, 15, 14, 13, 12, 11 };
shuffleArray(solutionArray);
for (int i = 0; i < solutionArray.length; i++)
{
System.out.print(solutionArray[i] + " ");
}
System.out.println();
}
// Implementing Fisher–Yates shuffle
static void shuffleArray(int[] ar)
{
// If running on Java 6 or older, use `new Random()` on RHS here
Random rnd = ThreadLocalRandom.current();
for (int i = ar.length - 1; i > 0; i--)
{
int index = rnd.nextInt(i + 1);
// Simple swap
int a = ar[index];
ar[index] = ar[i];
ar[i] = a;
}
}
}
这是另一种洗牌的方法
public List<Integer> shuffleArray(List<Integer> a) {
List<Integer> b = new ArrayList<Integer>();
while (a.size() != 0) {
int arrayIndex = (int) (Math.random() * (a.size()));
b.add(a.get(arrayIndex));
a.remove(a.get(arrayIndex));
}
return b;
}
从原始列表中选择一个随机数,并将其保存在另一个列表中。然后从原始列表中删除该数字。原始列表的大小将继续减小1,直到所有元素都移动到新列表中。
import java.util.ArrayList;
import java.util.Random;
public class shuffle {
public static void main(String[] args) {
int a[] = {1,2,3,4,5,6,7,8,9};
ArrayList b = new ArrayList();
int i=0,q=0;
Random rand = new Random();
while(a.length!=b.size())
{
int l = rand.nextInt(a.length);
//this is one option to that but has a flaw on 0
// if(a[l] !=0)
// {
// b.add(a[l]);
// a[l]=0;
//
// }
//
// this works for every no.
if(!(b.contains(a[l])))
{
b.add(a[l]);
}
}
// for (int j = 0; j <b.size(); j++) {
// System.out.println(b.get(j));
//
// }
System.out.println(b);
}
}
下面是一个使用数组列表的简单方法:
List<Integer> solution = new ArrayList<>();
for (int i = 1; i <= 6; i++) {
solution.add(i);
}
Collections.shuffle(solution);