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Class ArrayKShift

src/main/java/com/crossoverjie/algorithm/ArrayKShift.java:16–75  ·  view source on GitHub ↗

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

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14 * @since 1.0
15 */
16public 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

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