1 /**
  2      * Sorts the specified range of the array using the given
  3      * workspace array slice if possible for merging
  4      *
  5      * @param a the array to be sorted
  6      * @param left the index of the first element, inclusive, to be sorted
  7      * @param right the index of the last element, inclusive, to be sorted
  8      * @param work a workspace array (slice)
  9      * @param workBase origin of usable space in work array
 10      * @param workLen usable size of work array
 11      */
 12     static void sort(int[] a, int left, int right,
 13                      int[] work, int workBase, int workLen) {
 14         // Use Quicksort on small arrays
 15         if (right - left < QUICKSORT_THRESHOLD) {
 16             sort(a, left, right, true);
 17             return;
 18         }
 19 
 20         /*
 21          * Index run[i] is the start of i-th run
 22          * (ascending or descending sequence).
 23          */
 24         int[] run = new int[MAX_RUN_COUNT + 1];
 25         int count = 0; run[0] = left;
 26 
 27         // Check if the array is nearly sorted
 28         for (int k = left; k < right; run[count] = k) {
 29             if (a[k] < a[k + 1]) { // ascending
 30                 while (++k <= right && a[k - 1] <= a[k]);
 31             } else if (a[k] > a[k + 1]) { // descending
 32                 while (++k <= right && a[k - 1] >= a[k]);
 33                 for (int lo = run[count] - 1, hi = k; ++lo < --hi; ) {
 34                     int t = a[lo]; a[lo] = a[hi]; a[hi] = t;
 35                 }
 36             } else { // equal
 37                 for (int m = MAX_RUN_LENGTH; ++k <= right && a[k - 1] == a[k]; ) {
 38                     if (--m == 0) {
 39                         sort(a, left, right, true);
 40                         return;
 41                     }
 42                 }
 43             }
 44 
 45             /*
 46              * The array is not highly structured,
 47              * use Quicksort instead of merge sort.
 48              */
 49             if (++count == MAX_RUN_COUNT) {
 50                 sort(a, left, right, true);
 51                 return;
 52             }
 53         }
 54 
 55         // Check special cases
 56         // Implementation note: variable "right" is increased by 1.
 57         if (run[count] == right++) { // The last run contains one element
 58             run[++count] = right;
 59         } else if (count == 1) { // The array is already sorted
 60             return;
 61         }
 62 
 63         // Determine alternation base for merge
 64         byte odd = 0;
 65         for (int n = 1; (n <<= 1) < count; odd ^= 1);
 66 
 67         // Use or create temporary array b for merging
 68         int[] b;                 // temp array; alternates with a
 69         int ao, bo;              // array offsets from \'left\'
 70         int blen = right - left; // space needed for b
 71         if (work == null || workLen < blen || workBase + blen > work.length) {
 72             work = new int[blen];
 73             workBase = 0;
 74         }
 75         if (odd == 0) {
 76             System.arraycopy(a, left, work, workBase, blen);
 77             b = a;
 78             bo = 0;
 79             a = work;
 80             ao = workBase - left;
 81         } else {
 82             b = work;
 83             ao = 0;
 84             bo = workBase - left;
 85         }
 86 
 87         // Merging
 88         for (int last; count > 1; count = last) {
 89             for (int k = (last = 0) + 2; k <= count; k += 2) {
 90                 int hi = run[k], mi = run[k - 1];
 91                 for (int i = run[k - 2], p = i, q = mi; i < hi; ++i) {
 92                     if (q >= hi || p < mi && a[p + ao] <= a[q + ao]) {
 93                         b[i + bo] = a[p++ + ao];
 94                     } else {
 95                         b[i + bo] = a[q++ + ao];
 96                     }
 97                 }
 98                 run[++last] = hi;
 99             }
100             if ((count & 1) != 0) {
101                 for (int i = right, lo = run[count - 1]; --i >= lo;
102                     b[i + bo] = a[i + ao]
103                 );
104                 run[++last] = right;
105             }
106             int[] t = a; a = b; b = t;
107             int o = ao; ao = bo; bo = o;
108         }
109     }

 1 package recursion;
 2 
 3 import java.util.Arrays;
 4 
 5 /**
 6  * @author zsh
 7  * @company wlgzs
 8  * @create 2019-02-17 9:33
 9  * @Describe Arrays方法测试
10  */
11 public class TestForArrays {
12 
13     public static void main(String[] args) {
14         //填充数组，将arr[]中所有元素的值初始为0
15         int[] arr = new int[5];
16         Arrays.fill(arr,12);
17         System.out.println(Arrays.toString(arr));
18         //将arr中的第2个到第三个元素的值赋为8
19         Arrays.fill(arr,1,3,8);
20         System.out.println(Arrays.toString(arr));
21         //对数组进行排序
22         int[] arr1 = new int[]{7,6,8,5,2,9,8,1,3,5};
23         //对数组的第二个到第6个元素进行排序
24         Arrays.sort(arr1,1,6);
25         System.out.println(Arrays.toString(arr1));
26         //对整个数组进行排序
27         Arrays.sort(arr1);
28         System.out.println(Arrays.toString(arr1));
29         //比较数组元素是否相等
30         System.out.println(Arrays.equals(arr,arr1));
31         //使用二分法在数组中查找指定元素所在的下标
32         //数组必须是先排好序的
33         System.out.println(Arrays.binarySearch(arr1,5));
34         //如果不存在，就返回负数
35         System.out.println(Arrays.binarySearch(arr1,20));
36     }
37 
38 
39 }