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

使用Guava的int . aslist()就像这样简单:

Collections.shuffle(Ints.asList(array));

其他回答

还有另一种方法,还没有发布

//that way, send many object types diferentes
public anotherWayToReciveParameter(Object... objects)
{
    //ready with array
    final int length =objects.length;
    System.out.println(length);
    //for ready same list
    Arrays.asList(objects);
}

这种方法更简单,取决于上下文

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);

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]

你现在可以使用java 8了:

Collections.addAll(list, arr);
Collections.shuffle(list);
cardsList.toArray(arr);

最简单的解决方案,这种随机洗牌数组。

String location[] = {"delhi","banglore","mathura","lucknow","chandigarh","mumbai"};
int index;
String temp;
Random random = new Random();
for(int i=1;i<location.length;i++)
{
    index = random.nextInt(i+1);
    temp = location[index];
    location[index] = location[i];
    location[i] = temp;
    System.out.println("Location Based On Random Values :"+location[i]);
}