/**

      * Sorts the specified range of the array using the given

      * workspace array slice if possible for merging

      *

      * @param a the array to be sorted

      * @param left the index of the first element, inclusive, to be sorted

      * @param right the index of the last element, inclusive, to be sorted

      * @param work a workspace array (slice)

      * @param workBase origin of usable space in work array

      * @param workLen usable size of work array

      */

     static void sort(int[] a, int left, int right,

                      int[] work, int workBase, int workLen) {

         // Use Quicksort on small arrays

         if (right - left < QUICKSORT_THRESHOLD) {

             sort(a, left, right, true);

             return;

         }

         /*

          * Index run[i] is the start of i-th run

          * (ascending or descending sequence).

          */

         int[] run = new int[MAX_RUN_COUNT + 1];

         int count = 0; run[0] = left;

         // Check if the array is nearly sorted

         for (int k = left; k < right; run[count] = k) {

             if (a[k] < a[k + 1]) { // ascending

                 while (++k <= right && a[k - 1] <= a[k]);

             } else if (a[k] > a[k + 1]) { // descending

                 while (++k <= right && a[k - 1] >= a[k]);

                 for (int lo = run[count] - 1, hi = k; ++lo < --hi; ) {

                     int t = a[lo]; a[lo] = a[hi]; a[hi] = t;

                 }

             } else { // equal

                 for (int m = MAX_RUN_LENGTH; ++k <= right && a[k - 1] == a[k]; ) {

                     if (--m == 0) {

                         sort(a, left, right, true);

                         return;

                     }

                 }

             }

             /*

              * The array is not highly structured,

              * use Quicksort instead of merge sort.

              */

             if (++count == MAX_RUN_COUNT) {

                 sort(a, left, right, true);

                 return;

             }

         }

         // Check special cases

         // Implementation note: variable "right" is increased by 1.

         if (run[count] == right++) { // The last run contains one element

             run[++count] = right;

         } else if (count == 1) { // The array is already sorted

             return;

         }

         // Determine alternation base for merge

         byte odd = 0;

         for (int n = 1; (n <<= 1) < count; odd ^= 1);

         // Use or create temporary array b for merging

         int[] b;                 // temp array; alternates with a

         int ao, bo;              // array offsets from 'left'

         int blen = right - left; // space needed for b

         if (work == null || workLen < blen || workBase + blen > work.length) {

             work = new int[blen];

             workBase = 0;

         }

         if (odd == 0) {

             System.arraycopy(a, left, work, workBase, blen);

             b = a;

             bo = 0;

             a = work;

             ao = workBase - left;

         } else {

             b = work;

             ao = 0;

             bo = workBase - left;

         }

         // Merging

         for (int last; count > 1; count = last) {

             for (int k = (last = 0) + 2; k <= count; k += 2) {

                 int hi = run[k], mi = run[k - 1];

                 for (int i = run[k - 2], p = i, q = mi; i < hi; ++i) {

                     if (q >= hi || p < mi && a[p + ao] <= a[q + ao]) {

                         b[i + bo] = a[p++ + ao];

                     } else {

                         b[i + bo] = a[q++ + ao];

                     }

                 }

                 run[++last] = hi;

             }

             if ((count & 1) != 0) {

                 for (int i = right, lo = run[count - 1]; --i >= lo;

                     b[i + bo] = a[i + ao]

                 );

                 run[++last] = right;

             }

             int[] t = a; a = b; b = t;

             int o = ao; ao = bo; bo = o;

         }

     }

 package recursion;

 import java.util.Arrays;

 /**

  * @author zsh

  * @company wlgzs

  * @create 2019-02-17 9:33

  * @Describe Arrays方法测试

  */

 public class TestForArrays {

     public static void main(String[] args) {

         //填充数组，将arr[]中所有元素的值初始为0

         int[] arr = new int[5];

         Arrays.fill(arr,12);

         System.out.println(Arrays.toString(arr));

         //将arr中的第2个到第三个元素的值赋为8

         Arrays.fill(arr,1,3,8);

         System.out.println(Arrays.toString(arr));

         //对数组进行排序

         int[] arr1 = new int[]{7,6,8,5,2,9,8,1,3,5};

         //对数组的第二个到第6个元素进行排序

         Arrays.sort(arr1,1,6);

         System.out.println(Arrays.toString(arr1));

         //对整个数组进行排序

         Arrays.sort(arr1);

         System.out.println(Arrays.toString(arr1));

         //比较数组元素是否相等

         System.out.println(Arrays.equals(arr,arr1));

         //使用二分法在数组中查找指定元素所在的下标

         //数组必须是先排好序的

         System.out.println(Arrays.binarySearch(arr1,5));

         //如果不存在，就返回负数

         System.out.println(Arrays.binarySearch(arr1,20));

     }

 }