具有o(n)个时间复杂度和o(1)个空间复杂度的解。

void reverse(int[] array) {
    int start = 0;
    int end = array.length - 1;
    while (start < end) {
        int temp = array[start];
        array[start] = array[end];
        array[end] = temp;
        start++;
        end--;
    }
}

您的程序只适用于length = 0,1。 你可以试试:

int i = 0, j = validData.length-1 ; 
while(i < j)
{
     swap(validData, i++, j--);  // code for swap not shown, but easy enough
}

对非基本类型数组使用泛型的实现。

    //Reverse and get new Array -preferred
    public static final <T> T[] reverse(final T[] array) {
        final int len = array.length;
        final T[] reverse = (T[]) Array.newInstance(array.getClass().getComponentType(), len);
        for (int i = 0; i < len; i++) {
            reverse[i] = array[len-(i+1)];
        }
        return reverse;
    }
    
    //Reverse existing array - don't have to return it
    public static final <T> T[] reverseExisting(final T[] array) {
        final int len = array.length;
        for (int i = 0; i < len/2; i++) {
            final T temp = array[i];
            array[i] = array[len-(i+1)];
            array[len-(i+1)] = temp;
        }
        return array;
    }

番石榴

使用谷歌Guava库:

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

试试这个程序在JAVA:-

import java.util.Scanner;

public class Rev_one_D {

    static int row;

    static int[] trans_arr = new int[row];

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        row = n;

        int[] arr = new int[row];
        for (int i = 0; i < row; i++) {

            arr[i] = sc.nextInt();
            System.out.print(arr[i] + " ");

            System.out.println();
        }

        for (int i = 0; i < arr.length / 2; i++) {
            int temp = arr[i];
            arr[i] = arr[arr.length - i - 1];
            arr[arr.length - i - 1] = temp;

        }

        for (int i = 0; i < row; i++) {
            System.out.print(arr[i] + " ");
            System.out.println();
        }
    }
}

在Java 8的情况下，我们还可以使用IntStream来反转整数数组:

int[] sample = new int[]{1,2,3,4,5};
int size = sample.length;
int[] reverseSample = IntStream.range(0,size).map(i -> sample[size-i-1])
                      .toArray(); //Output: [5, 4, 3, 2, 1]