最有效的方法是向后迭代数组。

我不确定Aaron的解决方案是否调用Collections.reverse(list);有人知道吗?

简单的方法:

    for(int i=queue.length-1;i>=0;i--){
                System.out.print(queue[i] + "  ");
    }

以下是我想到的:

// solution 1 - boiler plated 
Integer[] original = {100, 200, 300, 400};
Integer[] reverse = new Integer[original.length];

int lastIdx = original.length -1;
int startIdx = 0;

for (int endIdx = lastIdx; endIdx >= 0; endIdx--, startIdx++)
   reverse[startIdx] = original[endIdx];

System.out.printf("reverse form: %s", Arrays.toString(reverse));

// solution 2 - abstracted 
// convert to list then use Collections static reverse()
List<Integer> l = Arrays.asList(original);
Collections.reverse(l);
System.out.printf("reverse form: %s", l);

使用流来反转

这里已经有很多答案，主要集中在就地修改数组。但是为了完整起见，这里有另一种使用Java流来保留原始数组并创建一个新的反向数组的方法:

int[] a = {8, 6, 7, 5, 3, 0, 9};
int[] b = IntStream.rangeClosed(1, a.length).map(i -> a[a.length-i]).toArray();

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();
        }
    }
}

static int[] reverseArray(int[] a) {
     int ret[] = new int[a.length];
     for(int i=0, j=a.length-1; i<a.length && j>=0; i++, j--)
         ret[i] = a[j];
     return ret;
}