数组右移K次, 原数组 [1, 2, 3, 4, 5, 6, 7] 右移3次后结果为 [5,6,7,1,2,3,4] 基本思路:不开辟新的数组空间的情况下考虑在原属组上进行操作 1 将数组倒置,这样后k个元素就跑到了数组的前面,然后反转一下即可 2 同理后 len-k个元素只需要翻转就完成数组的k次移动 @author 656369960@qq.com @date 12/7/2018 1:38 PM @since 1.0
| 14 | * @since 1.0 |
| 15 | */ |
| 16 | public class ArrayKShift { |
| 17 | |
| 18 | public void arrayKShift(int[] array, int k) { |
| 19 | |
| 20 | /** |
| 21 | * constrictions |
| 22 | */ |
| 23 | |
| 24 | if (array == null || 0 == array.length) { |
| 25 | return ; |
| 26 | } |
| 27 | |
| 28 | k = k % array.length; |
| 29 | |
| 30 | if (0 > k) { |
| 31 | return; |
| 32 | } |
| 33 | |
| 34 | |
| 35 | /** |
| 36 | * reverse array , e.g: [1, 2, 3 ,4] to [4,3,2,1] |
| 37 | */ |
| 38 | |
| 39 | for (int i = 0; i < array.length / 2; i++) { |
| 40 | int tmp = array[i]; |
| 41 | array[i] = array[array.length - 1 - i]; |
| 42 | array[array.length - 1 - i] = tmp; |
| 43 | } |
| 44 | |
| 45 | /** |
| 46 | * first k element reverse |
| 47 | */ |
| 48 | for (int i = 0; i < k / 2; i++) { |
| 49 | int tmp = array[i]; |
| 50 | array[i] = array[k - 1 - i]; |
| 51 | array[k - 1 - i] = tmp; |
| 52 | } |
| 53 | |
| 54 | /** |
| 55 | * last length - k element reverse |
| 56 | */ |
| 57 | |
| 58 | for (int i = k; i < k + (array.length - k ) / 2; i ++) { |
| 59 | int tmp = array[i]; |
| 60 | array[i] = array[array.length - 1 - i + k]; |
| 61 | array[array.length - 1 - i + k] = tmp; |
| 62 | } |
| 63 | } |
| 64 | |
| 65 | public static void main(String[] args) { |
| 66 | int[] array = {1, 2, 3 ,4, 5, 6, 7}; |
| 67 | ArrayKShift shift = new ArrayKShift(); |
| 68 | shift.arrayKShift(array, 6); |
| 69 | |
| 70 | Arrays.stream(array).forEach(o -> { |
| 71 | System.out.println(o); |
| 72 | }); |
| 73 |
nothing calls this directly
no outgoing calls
no test coverage detected