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
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]
其他回答
Groovy的一个简单解决方案:
solutionArray.sort{ new Random().nextInt() }
这将对数组列表中的所有元素进行随机排序,存档所有元素洗牌的预期结果。
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;
}
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]
你应该使用Collections.shuffle()。但是,不能直接操作原始类型数组,因此需要创建包装器类。
试试这个。
public static void shuffle(int[] array) {
Collections.shuffle(new AbstractList<Integer>() {
@Override public Integer get(int index) { return array[index]; }
@Override public int size() { return array.length; }
@Override public Integer set(int index, Integer element) {
int result = array[index];
array[index] = element;
return result;
}
});
}
And
int[] solutionArray = {1, 2, 3, 4, 5, 6, 6, 5, 4, 3, 2, 1};
shuffle(solutionArray);
System.out.println(Arrays.toString(solutionArray));
输出:
[3, 3, 4, 1, 6, 2, 2, 1, 5, 6, 5, 4]
Random rnd = new Random();
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;
}
顺便说一下,我注意到这段代码返回了一个ar.length - 1个数的元素,所以如果你的数组有5个元素,新的打乱的数组将有4个元素。这是因为for循环说i>0。如果你改变为i>=0,你得到所有的元素洗牌。