我有一个数组:
int[] arr1 = {1, 2, 3, 4, 5, 6, 7, 8, 9};我想将数组从指定范围旋转 k 倍，即从第五个元素到最后一个元素:
And I want to rotate the array from a specified range by k times i.e. from 5th element to last element:
output: {1, 2, 3, 4, 9, 8, 5, 6, 7}我试图使算法从数组的开始到结尾旋转
I tried to adapt the algorithm to rotate from start to end of the array
public static int[] rotate(int[] nums, int k) { int[] a = new int[nums.length]; for (int i = 4; i < nums.length; i++) { a[(i + k) % nums.length] = nums[i]; } for (int i = 4; i < nums.length; i++) { nums[i] = a[i + 4]; } return nums; }但是，输出为: 1 2 3 4 0 0 5 5 7
除了将目标元素从原始数组复制到临时数组中然后运行此算法之外，我在做什么错?为什么返回 0，0 而不是 9，8 ?
Apart from copying the target elements from the original array into a temporary array and then running this algorithm, what am I doing wrong? Why is 0, 0 returned instead of 9, 8?
推荐答案按指定步骤在指定范围之间旋转数组中的元素:
Rotate an elements in an array between a specified range by a specified step:
将数组分为三个部分:之前，范围和之后.
移动指定范围的数组.
2.1.将范围分为两部分: near 和 far .
2.1. Split this range into two parts: near and far.
2.2.交换它们并串联起来.
2.2. Swap them and concatenate back.
将所有内容重新连接起来.
Concatenate everything back.
public static void main(String[] args) { int[] arr = {1, 2, 3, 4, 5, 6, 7, 8, 9}; int[] rotated = rotateRange(arr, 4, arr.length, 3); System.out.println(Arrays.toString(rotated)); // [1, 2, 3, 4, 8, 9, 5, 6, 7] } // rotate a specified range of an array by a specified step static int[] rotateRange(int[] arr, int start, int end, int n) { return Stream.of( // three parts: 'before', 'range' and 'after' Arrays.stream(arr, 0, start), // get a specified range and rotate it by a specified step Arrays.stream(rotate(Arrays.copyOfRange(arr, start, end), n)), Arrays.stream(arr, end, arr.length)) // flatten into one stream .flatMapToInt(Function.identity()) // return an array .toArray(); } // rotate an array by a specified step static int[] rotate(int[] arr, int n) { // prevent circular rotation n = n % arr.length; return IntStream.concat( // concatenate the two parts: 'far' and 'near' Arrays.stream(arr, n, arr.length), Arrays.stream(arr, 0, n)) // return an array .toArray(); }另请参见:仅使用一个分号在Java中旋转int数组