关键词：
Arrays排序算法
import java.util.Arrays;
排序算法 - 数据结构
Arrays.sort耗时
100000 个数升序排列耗时测试:
耗时: 14、耗时: 10、耗时: 14、耗时: 11、耗时: 13、耗时: 10、耗时: 11、耗时: 12、耗时: 17、耗时: 15
从时间上查看排序算法 - 数据结构，与快速排序、堆排序、计数排序接近。
双基准快速排序
DualPivotQuicksort.sort
这里参数中right是数组a.length - 1
使用快速排序
数组小于286 使用快速排序
private static final int QUICKSORT_THRESHOLD = 286;
优先使用插入排序
数组小于47 则优先使用插入排序而不是快速排序
private static final int INSERTION_SORT_THRESHOLD = 47;
/*
* 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;
Arrays.sort中快速排序
private static void sort(int[] a, int left, int right, boolean leftmost)
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; )
int ak = 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);
>QUICKSORT_THRESHOLD 情况
- 数组不是高度结构化的，请使用快速排序而不是合并排序。
- 使用或创建临时数组b进行合并
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)
// Equal items in the beginning of the sequence
while (k < right && a[k] == a[k + 1])
k++;
if (k == right) break; // Sequence finishes with equal items
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]);
// Transform into an ascending sequence
for (int lo = run[count] - 1, hi = k; ++lo < --hi; )
int t = a[lo]; a[lo] = a[hi]; a[hi] = t;
// Merge a transformed descending sequence followed by an
// ascending sequence
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