在Java中反转数组[duplicate]

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• How do I reverse an int array in Java? 38 answers
• 如何在Java中反转int数组?38个答案

If I have an array like this:

如果我有一个这样的数组:

1 4 9 16 9 7 4 9 11 

What is the best way to reverse the array so that it looks like this:

使数组看起来像这样的最好方法是什么?

11 9 4 7 9 16 9 4 1 

I have the code below, but I feel it is a little tedious:

我有下面的代码，但我觉得有点乏味:

public int[] reverse3(int[] nums) {
    return new int[] { nums[8], nums[7], nums[6], nums[5], num[4],
                       nums[3], nums[2], nums[1], nums[0] };
}

Is there a simpler way?

有更简单的方法吗?

15 个解决方案

#1

Collections.reverse() can do that job for you if you put your numbers in a List of Integers.

如果您将数字放在整数列表中，那么reverse()可以为您完成这项工作。

List<Integer> list = Arrays.asList(1, 4, 9, 16, 9, 7, 4, 9, 11);
System.out.println(list);
Collections.reverse(list);
System.out.println(list);

Output:

输出:

[1, 4, 9, 16, 9, 7, 4, 9, 11]
[11, 9, 4, 7, 9, 16, 9, 4, 1]

#2

If you want to reverse the array in-place:

如果你想反转数组的位置:

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

It works since Arrays.asList returns a write-through proxy to the original array.

它工作以来数组。asList返回一个写入代理到原始数组。

#3

If you don't want to use Collections then you can do this:

如果你不想使用集合，你可以这样做:

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

#4

I like to keep the original array and return a copy. This is a generic version:

我喜欢保留原始数组并返回一个副本。这是一个通用版本:

public static <T> T[] reverse(T[] array) {
    T[] copy = array.clone();
    Collections.reverse(Arrays.asList(copy));
    return copy;
}

without keeping the original array:

不保留原数组:

public static <T> void reverse(T[] array) {
    Collections.reverse(Arrays.asList(array));
}

#5

try this:

试试这个:

public int[] reverse3(int[] nums) {
    int[] reversed = new int[nums.length];
    for (int i=0; i<nums.length; i++) {
        reversed[i] = nums[nums.length - 1 - i];
    }
    return reversed;
}

My input was:

我输入的是:

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12

12 3 4 5 6 7 8 9 10 11 12

And the output I got:

我得到的输出是

12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1

12 11 10 9 8 7 6 5 4 3 2 1

#6

You could use org.apache.commons.lang.ArrayUtils : ArrayUtils.reverse(array)

你可以使用org.apache.commons.lang.ArrayUtils: ArrayUtils.reverse(array)

#7

In case you don't want to use a temporary variable, you can also do like this:

如果您不想使用临时变量，也可以这样做:

final int len = arr.length;
for (int i=0; i < (len/2); i++) {
    arr[i] += arr[len - 1 - i]; //  a = a+b
    arr[len - 1 - i] = arr[i] - arr[len - 1 - i];   //  b = a-b
    arr[i] -= arr[len - 1 - i]; //  a = a-b
}

#8

In place reversal with minimum amount of swaps.

用最少的掉期来代替。

for (int i = 0; i < a.length / 2; i++) {
    int tmp = a[i];
    a[i] = a[a.length - 1 - i];
    a[a.length - 1 - i] = tmp;
}

#9

Or you could loop through it backeards

或者你可以把它圈起来。

int[] firstArray = new int[]{1,2,3,4};
int[] reversedArray = new int[firstArray.length];
int j = 0;
for (int i = firstArray.length -1; i > 0; i--){
    reversedArray[j++] = firstArray[i];
}

(note: I have not compiled this but hopefully it is correct)

(注意:我还没有编译它，但希望它是正确的)

#10

I would do something like this:

我会这样做:

public int[] reverse3(int[] nums) {
  int[] numsReturn = new int[nums.length()]; 
  int count = nums.length()-1;
  for(int num : nums) {
    numsReturn[count] = num;
    count--;
  }
  return numsReturn;
}

#11

you messed up

你搞砸了

int[] firstArray = new int[]{1,2,3,4};
int[] reversedArray = new int[firstArray.length];
int j = 0;
for (int i = firstArray.length -1; i >= 0; i--){
    reversedArray[j++] = firstArray[i];
}

#12

 public void swap(int[] arr,int a,int b)
 {
    int temp=arr[a];
    arr[a]=arr[b];
    arr[b]=temp;        
}
public int[] reverseArray(int[] arr){
    int size=arr.length-1;

    for(int i=0;i<size;i++){

        swap(arr,i,size--); 

    }

    return arr;
}

#13

The following will reverse in place the array between indexes i and j (to reverse the whole array call reverse(a, 0, a.length - 1))

下面将在索引i和j之间反向放置数组(反向反转整个数组调用)。长度- 1))

    public void reverse(int[] a, int i , int j) {
        int ii =  i;
        int jj = j;

        while (ii < jj) {
            swap(ii, jj);
            ++ii;
            --jj;
        }
    }

#14

This code would help:

这段代码将帮助:

int [] a={1,2,3,4,5,6,7};
for(int i=a.length-1;i>=0;i--)
  System.out.println(a[i]);

#15

you can send the original array to a method for example:

您可以将原始数组发送到一个方法，例如:

after that you create a new array to hold the reversed elements

在此之后，您将创建一个新的数组来保存反转的元素。

public static void reverse(int[] a){

int[] reversedArray = new int[a.length];

for(int i = 0 ; i<a.length; i++){
reversedArray[i] = a[a.length -1 -i];


}