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
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);
}
}
其他回答
使用随机类
public static void randomizeArray(int[] arr) {
Random rGenerator = new Random(); // Create an instance of the random class
for (int i =0; i< arr.length;i++ ) {
//Swap the positions...
int rPosition = rGenerator.nextInt(arr.length); // Generates an integer within the range (Any number from 0 - arr.length)
int temp = arr[i]; // variable temp saves the value of the current array index;
arr[i] = arr[rPosition]; // array at the current position (i) get the value of the random generated
arr[rPosition] = temp; // the array at the position of random generated gets the value of temp
}
for(int i = 0; i<arr.length; i++) {
System.out.print(arr[i]); //Prints out the array
}
}
类似的情况没有使用swap b
Random r = new Random();
int n = solutionArray.length;
List<Integer> arr = Arrays.stream(solutionArray)
.boxed()
.collect(Collectors.toList());
for (int i = 0; i < n-1; i++) {
solutionArray[i] = arr.remove(r.nextInt(arr.size())); // randomize based on size
}
solutionArray[n-1] = arr.get(0);
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]
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;
}
}
}
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));
}