我试图在Java中反转一个int数组。

此方法不反转数组。

for(int i = 0; i < validData.length; i++)
{
    int temp = validData[i];
    validData[i] = validData[validData.length - i - 1];
    validData[validData.length - i - 1] = temp;
}

有什么问题吗?


当前回答

简单for循环!

for (int start = 0, end = array.length - 1; start <= end; start++, end--) {
    int aux = array[start];
    array[start]=array[end];
    array[end]=aux;
}

其他回答

这里是一个简单的实现,反转数组的任何类型,加上全/部分支持。

import java.util.logging.Logger;

public final class ArrayReverser {
 private static final Logger LOGGER = Logger.getLogger(ArrayReverser.class.getName());

 private ArrayReverser () {

 }

 public static <T> void reverse(T[] seed) {
    reverse(seed, 0, seed.length);
 }

 public static <T> void reverse(T[] seed, int startIndexInclusive, int endIndexExclusive) {
    if (seed == null || seed.length == 0) {
        LOGGER.warning("Nothing to rotate");
    }
    int start = startIndexInclusive < 0 ? 0 : startIndexInclusive;
    int end = Math.min(seed.length, endIndexExclusive) - 1;
    while (start < end) {
        swap(seed, start, end);
        start++;
        end--;
    }
}

 private static <T> void swap(T[] seed, int start, int end) {
    T temp =  seed[start];
    seed[start] = seed[end];
    seed[end] = temp;
 }  

}

下面是相应的单元测试

import static org.hamcrest.CoreMatchers.is;
import static org.junit.Assert.assertThat;

import org.junit.Before;
import org.junit.Test;

public class ArrayReverserTest {
private Integer[] seed;

@Before
public void doBeforeEachTestCase() {
    this.seed = new Integer[]{1,2,3,4,5,6,7,8};
}

@Test
public void wholeArrayReverse() {
    ArrayReverser.<Integer>reverse(seed);
    assertThat(seed[0], is(8));
}

 @Test
 public void partialArrayReverse() {
    ArrayReverser.<Integer>reverse(seed, 1, 5);
    assertThat(seed[1], is(5));
 }
}
Collections.reverse(Arrays.asList(yourArray));

java.util.Collections.reverse()可以反转java.util.Lists和java.util.Arrays.asList()返回一个列表,该列表包装了您传递给它的特定数组,因此在调用Collections.reverse()之后,yourArray将被反转。

其代价只是创建一个list对象,不需要额外的库。

在Tarik和他们的评论者的回答中已经提出了一个类似的解决方案,但我认为这个答案会更简洁,更容易被分析。

这是我个人的解决方法。创建参数化方法的原因是允许对任何数组进行排序…不仅仅是整数。

我希望你能从中有所收获。

@Test
public void reverseTest(){
   Integer[] ints = { 1, 2, 3, 4 };
   Integer[] reversedInts = reverse(ints);

   assert ints[0].equals(reversedInts[3]);
   assert ints[1].equals(reversedInts[2]);
   assert ints[2].equals(reversedInts[1]);
   assert ints[3].equals(reversedInts[0]);

   reverseInPlace(reversedInts);
   assert ints[0].equals(reversedInts[0]);
}

@SuppressWarnings("unchecked")
private static <T> T[] reverse(T[] array) {
    if (array == null) {
        return (T[]) new ArrayList<T>().toArray();
    }
    List<T> copyOfArray = Arrays.asList(Arrays.copyOf(array, array.length));
    Collections.reverse(copyOfArray);
    return copyOfArray.toArray(array);
}

private static <T> T[] reverseInPlace(T[] array) {
    if(array == null) {
        // didn't want two unchecked suppressions
        return reverse(array);
    }

    Collections.reverse(Arrays.asList(array));
    return array;
}
int[] arrTwo = {5, 8, 18, 6, 20, 50, 6};

    for (int i = arrTwo.length-1; i > 0; i--)
    {
        System.out.print(arrTwo[i] + " ");
    }
   import java.util.Scanner;
class ReverseArray 
{
    public static void main(String[] args) 
    {
        int[] arra = new int[10];
        Scanner sc = new Scanner(System.in);
        System.out.println("Enter Array Elements : ");
        for(int i = 0 ; i <arra.length;i++)
        {
            arra[i] = sc.nextInt();
        }

        System.out.println("Printing  Array : ");
        for(int i = 0; i <arra.length;i++)
        {
            System.out.print(arra[i] + " ");
        }

        System.out.println();
        System.out.println("Printing  Reverse Array : ");
        for(int i = arra.length-1; i >=0;i--)
        {
            System.out.print(arra[i] + " ");
        }
    }
}