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
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);
其他回答
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]);
}
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);
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);
}
}
下面是一个有效的Fisher-Yates shuffle数组函数:
private static void shuffleArray(int[] array)
{
int index;
Random random = new Random();
for (int i = array.length - 1; i > 0; i--)
{
index = random.nextInt(i + 1);
if (index != i)
{
array[index] ^= array[i];
array[i] ^= array[index];
array[index] ^= array[i];
}
}
}
or
private static void shuffleArray(int[] array)
{
int index, temp;
Random random = new Random();
for (int i = array.length - 1; i > 0; i--)
{
index = random.nextInt(i + 1);
temp = array[index];
array[index] = array[i];
array[i] = temp;
}
}
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));
}