最近找工作的时候总是会被问道一些简单的算法问题,其中最多就是将一个数组进行排序,每每我回答了冒泡排序,快速排序,二叉排序,都会再次问我还有没有更简单的方法,当时因为没有了解所以没有回答上来。之后经过查找,发现有一个叫做
Arrays.sort();  
的方法可以将数组以参数的形式传递进去,并且将数组从小到大进行排序,很尴尬的是在我再次回答这个问题的时候,面试官问我是否看过源码,我又是一脸懵逼,今天在这里记录一下源码的原理实现。

/** * Sorts the specified array into ascending numerical order. * * <p>
Implementation note: The sorting algorithm is a Dual-Pivot Quicksort * by
Vladimir Yaroslavskiy, Jon Bentley, and Joshua Bloch. This algorithm * offers
O(n log(n)) performance on many data sets that cause other * quicksorts to
degrade to quadratic performance, and is typically * faster than traditional
(one-pivot) Quicksort implementations. * * @param a the array to be sorted */
public static voidsort(int[] a) { DualPivotQuicksort.sort(a, 0, a.length - 1,
null,0, 0); }

这段注释的翻译

各种指定的数组升序顺序。
*
* <BR>实施注:排序算法是一种Dual Pivot Quicksort
* Vladimir Yaroslavskiy,Jon Bentley,Joshua Bloch。该算法
*提供了O(n log(n))的性能在许多数据集,引起其他
* quicksorts退化的二次型性能,通常是
*比传统(一个支点)快速排序的实现。

从这个方法可以看出,我们虽然只传入一个数组的参数,但是方法中需要运算的参数并不是只有这个数组。我们先分析一个这个调用的方法中是如何实现的。

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;
into = ao; ao = bo; bo = o; } }

这是这个方法的全部实现。现在我们就来分析一个这个原理。

第一个if中

if (right - left < QUICKSORT_THRESHOLD) { sort(a, left, right, true); return;
}

QUICKSORT_THRESHOLD = 286;

如果这个数组的长度是大于286的时候,将会不再走这个方法,而是会走下面的方法,这个我们先不进行深入的研究,毕竟一般进行排序的数组没有这么长。

/** * Sorts the specified range of the array by Dual-Pivot Quicksort. * *
@parama 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 leftmost indicates if this part is the
leftmost in the range */ private static void sort(int[] a, int left, int right,
booleanleftmost) { int length = right - left + 1; // Use insertion sort on tiny
arrays if (length < INSERTION_SORT_THRESHOLD) { if (leftmost) { /* *
Traditional (without sentinel) insertion sort, * optimized for server VM, is
used in case of * the leftmost part. */ for (int i = left, j = i; i < right; j
= ++i) {int ai = a[i + 1]; while (ai < a[j]) { a[j + 1] = a[j]; if (j-- ==
left) {break; } } a[j + 1] = ai; } } else { /* * Skip the longest ascending
sequence. */ do { if (left >= right) { return; } } while (a[++left] >= a[left -
1]); /* * Every element from adjoining part plays the role * of sentinel,
therefore this allows us to avoid the * left range check on each iteration.
Moreover, we use * the more optimized algorithm, so called pair insertion *
sort, which is faster (in the context of Quicksort) * than traditional
implementation of insertion sort. */ for (int k = left; ++left <= right; k =
++left) {int a1 = a[k], a2 = a[left]; if (a1 < a2) { a2 = a1; a1 = a[left]; }
while(a1 < a[--k]) { a[k + 2] = a[k]; } a[++k + 1] = a1; while (a2 < a[--k]) {
a[k +1] = a[k]; } a[k + 1] = a2; } int last = a[right]; while (last <
a[--right]) { a[right +1] = a[right]; } a[right + 1] = last; } return; } //
Inexpensive approximation of length / 7 int seventh = (length >> 3) + (length >>
6) + 1; /* * Sort five evenly spaced elements around (and including) the *
center element in the range. These elements will be used for * pivot selection
as described below. The choice for spacing * these elements was empirically
determined to work well on * a wide variety of inputs. */ int e3 = (left +
right) >>>1; // The midpoint int e2 = e3 - seventh; int e1 = e2 - seventh; int
e4 = e3 + seventh; int e5 = e4 + seventh; // Sort these elements using
insertion sort if (a[e2] < a[e1]) { int t = a[e2]; a[e2] = a[e1]; a[e1] = t; }
if(a[e3] < a[e2]) { int t = a[e3]; a[e3] = a[e2]; a[e2] = t; if (t < a[e1]) {
a[e2] = a[e1]; a[e1] = t; } } if (a[e4] < a[e3]) { int t = a[e4]; a[e4] = a[e3];
a[e3] = t; if (t < a[e2]) { a[e3] = a[e2]; a[e2] = t; if (t < a[e1]) { a[e2]
= a[e1]; a[e1] = t; } } } if (a[e5] < a[e4]) { int t = a[e5]; a[e5] = a[e4];
a[e4] = t; if (t < a[e3]) { a[e4] = a[e3]; a[e3] = t; if (t < a[e2]) { a[e3]
= a[e2]; a[e2] = t; if (t < a[e1]) { a[e2] = a[e1]; a[e1] = t; } } } } //
Pointers int less = left; // The index of the first element of center part int
great = right; // The index before the first element of right part if (a[e1] !=
a[e2] && a[e2] != a[e3] && a[e3] != a[e4] && a[e4] != a[e5]) {/* * Use the
second and fourth of the five sorted elements as pivots. * These values are
inexpensive approximations of the first and * second terciles of the array.
Note that pivot1 <= pivot2. */ int pivot1 = a[e2]; int pivot2 = a[e4]; /* *
The first and the last elements to be sorted are moved to the * locations
formerly occupied by the pivots. When partitioning * is complete, the pivots
are swapped back into their final * positions, and excluded from subsequent
sorting. */ a[e2] = a[left]; a[e4] = a[right]; /* * Skip elements, which are
less or greater than pivot values. */ while (a[++less] < pivot1); while
(a[--great] > pivot2); /* * Partitioning: * * left part center part right
part * +--------------------------------------------------------------+ * | <
pivot1 | pivot1 <= && <= pivot2 | ? | > pivot2 | *
+--------------------------------------------------------------+ * ^ ^ ^ * | |
| * less k great * * Invariants: * * all in (left, less) < pivot1 * pivot1
<= all in [less, k) <= pivot2 * all in (great, right) > pivot2 * * Pointer k
is the first index of ?-part. */ outer: for (int k = less - 1; ++k <= great; ) {
intak = a[k]; if (ak < pivot1) { // Move a[k] to left part a[k] = a[less]; /*
* Here and below we use "a[i] = b; i++;" instead * of "a[i++] = b;" due to
performance issue. */ a[less] = ak; ++less; } else if (ak > pivot2) { // Move
a[k] to right part while (a[great] > pivot2) { if (great-- == k) { break outer;
} }if (a[great] < pivot1) { // a[great] <= pivot2 a[k] = a[less]; a[less] =
a[great]; ++less; } else { // pivot1 <= a[great] <= pivot2 a[k] = a[great]; } /*
* Here and below we use "a[i] = b; i--;" instead * of "a[i--] = b;" due to
performance issue. */ a[great] = ak; --great; } } // Swap pivots into their
final positions a[left] = a[less - 1]; a[less - 1] = pivot1; a[right] = a[great
+1]; a[great + 1] = pivot2; // Sort left and right parts recursively, excluding
known pivots sort(a, left, less - 2, leftmost); sort(a, great + 2, right, false)
; /* * If center part is too large (comprises > 4/7 of the array), * swap
internal pivot values to ends. */ if (less < e1 && e5 < great) { /* * Skip
elements, which are equal to pivot values. */ while (a[less] == pivot1) { ++less
; } while (a[great] == pivot2) { --great; } /* * Partitioning: * * left part
center part right part *
+----------------------------------------------------------+ * | == pivot1 |
pivot1 < && < pivot2 | ? | == pivot2 | *
+----------------------------------------------------------+ * ^ ^ ^ * | | |
* less k great * * Invariants: * * all in (*, less) == pivot1 * pivot1 <
all in [less, k) < pivot2 * all in (great, *) == pivot2 * * Pointer k is the
first index of ?-part. */ outer: for (int k = less - 1; ++k <= great; ) { int
ak = a[k]; if (ak == pivot1) { // Move a[k] to left part a[k] = a[less];
a[less] = ak; ++less; } else if (ak == pivot2) { // Move a[k] to right part
while(a[great] == pivot2) { if (great-- == k) { break outer; } } if (a[great]
== pivot1) {// a[great] < pivot2 a[k] = a[less]; /* * Even though a[great]
equals to pivot1, the * assignment a[less] = pivot1 may be incorrect, * if
a[great] and pivot1 are floating-point zeros * of different signs. Therefore in
float and * double sorting methods we have to use more * accurate assignment
a[less] = a[great]. */ a[less] = pivot1; ++less; } else { // pivot1 < a[great]
< pivot2 a[k] = a[great]; } a[great] = ak; --great; } } } // Sort center part
recursively sort(a, less, great, false); } else { // Partitioning with one pivot
/* * Use the third of the five sorted elements as pivot. * This value is
inexpensive approximation of the median. */ int pivot = a[e3]; /* *
Partitioning degenerates to the traditional 3-way * (or "Dutch National Flag")
schema: * * left part center part right part *
+-------------------------------------------------+ * | < pivot | == pivot | ?
| > pivot | * +-------------------------------------------------+ * ^ ^ ^ * |
| | * less k great * * Invariants: * * all in (left, less) < pivot * all
in [less, k) == pivot * all in (great, right) > pivot * * Pointer k is the
first index of ?-part. */ for (int k = less; k <= great; ++k) { if (a[k] ==
pivot) {continue; } int ak = a[k]; if (ak < pivot) { // Move a[k] to left part
a[k] = a[less]; a[less] = ak; ++less; } else { // a[k] > pivot - Move a[k] to
right part while (a[great] > pivot) { --great; } if (a[great] < pivot) { //
a[great] <= pivot a[k] = a[less]; a[less] = a[great]; ++less; } else { //
a[great] == pivot /* * Even though a[great] equals to pivot, the * assignment
a[k] = pivot may be incorrect, * if a[great] and pivot are floating-point *
zeros of different signs. Therefore in float * and double sorting methods we
have to use * more accurate assignment a[k] = a[great]. */ a[k] = pivot; }
a[great] = ak; --great; } } /* * Sort left and right parts recursively. * All
elements from center part are equal * and, therefore, already sorted. */ sort(a
,left, less - 1, leftmost); sort(a, great + 1, right, false); } }

首先这个方法中使用传入的俩个参数来对传入的数组进行遍历,如果前一个数小于后一个数

未完待续

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