DualPivotQuicksort源码解析
我们可以直接调用Arrays.sort()来排序一个数组，跟踪下去实际上调用的是DualPivotQuicksort中的sort（a,left,rigth,null,0,0）
int sort（对int类型的数组进行的排序）
/**
* 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
*/
//private static final int QUICKSORT_THRESHOLD = 286;
//整形排序：如果排序长度小于286,这直接进入双轴快排
//若大于则先判断数组是否大致有序，如果数组中有序段个数小于67个，这进行归并（合并）排序
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).
*/
//有序段的开始下标放在run数组里，第0位为left
//例如[1,2,3,1,2,1],那么run[1]为3，run[2]为5
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.
*/
//如果有序段的个数大于67个，进行双轴快排
if (++count == MAX_RUN_COUNT) {
sort(a, left, right, true);
return;
}
}
// Check special cases
// Implementation note: variable "right" is increased by 1.
//run数组的最后两个元素为right，right + 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
//判断归并排序的次数为奇数次还是偶数次，odd为0代表奇数次，odd为1代表偶数次
byte odd = 0;
for (int n = 1; (n <<= 1) < count; odd ^= 1);
// Use or create temporary array b for merging
//初始化辅助空间
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;
}
//若为奇数次，则先交换原数组和辅助数组，即a和b
//因为每一次归并排序的最后都需要再交换一次数组，若最终交换次数为奇数次，则提前交换一次可
//以保证a指向的是原数组
//至于ao，和bo是a和b两个数组互相映射的偏移量
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
//归并排序(这一段个人感觉偏难，但是可以尝试找一个具体的数组变量跟踪，感叹源码的牛b，自己
//感觉看懂也写不出来)
for (int last; count > 1; count = last) {
//p和q分别指向两个要合并的相邻的数组，hi表示第三个有序段的开始，然后进行两两合并
for (int k = (last = 0) + 2; k <= count; k += 2) {
int hi = run[k], mi = run[k - 1];
//将排序结果映射到b数组上
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];
}
}
//更新run数组
run[++last] = hi;
}
//若count为奇数，则表明还有最后一段没有排序，将最后一段复制到b数组上(或许用
//System.copy()更快，手动滑稽)
if ((count & 1) != 0) {
for (int i = right, lo = run[count - 1]; --i >= lo;
b[i + bo] = a[i + ao]
);
run[++last] = right;
}
//交换两个数组
int[] t = a; a = b; b = t;
int o = ao; ao = bo; bo = o;
}
}
/**
* Sorts the specified range of the array by Dual-Pivot Quicksort.
*
* @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 leftmost indicates if this part is the leftmost in the range
*/
//正式进入双轴快排
private static void sort(int[] a, int left, int right, boolean leftmost) {
int length = right - left + 1;
// Use insertion sort on tiny arrays
//若待排序个数小于47，则进行插入排序
if (length < INSERTION_SORT_THRESHOLD) {
//若leftmost为true则进行经典插入排序，否则进行成对插入排序
//leftmost按字面上的意思是最左端，但我并不理解为什么是最左端就适合经典插入
//有知道的小伙伴评论说明一下，谢谢
if (leftmost) {
/*
* Traditional (without sentinel) insertion sort,
* optimized for server VM, is used in case of
* the leftmost part.
*/
//经典插入排序，（比我自己写的插入排序好多了，手动哭）
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.
*/
//跳过有序的部分
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.
*/
//成对插入排序其实就是两个元素一起插排，这样更快吗（是不是可双轴快排一样）
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
//根据经验得出来的步长
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.
*/
//e3为中间元素，然后依次加减步长得出其它的e
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
//给所有的e进行暴力排序，e1最小，e5最大
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
//若满足下面这个条件，则以e2和e4进行双轴快排，否则以e3进行单轴快排
//也好理解，若非要找两个不一样的进行双轴是比较浪费时间的
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.
*/
//less前面的全是小于pivot1的，great后面的全是大于pivot2的，k指向正在遍历的数
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
//将less前的那一段，和great后的那一段进行递归
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.
*/
//若中间的那一段太大
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.
*/
//将所有和pivot1和pivot2相等的数移到两边
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
//对中间部分进行排序
sort(a, less, great, false);
} else {
//这个是单轴快排
// 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);
}
}
其它类型的数组的排序过程和int大致相同，我会将不同的地方一一列出
对long的排序完全一致
short sort
/**
* 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
*/
//若待排数组长度大于3200个，则进行计数排序
//因为short用两个字节表示所以最多表示65536个数字，因此用计数排序比较好
static void sort(short[] a, int left, int right,
short[] work, int workBase, int workLen) {
// Use counting sort on large arrays
if (right - left > COUNTING_SORT_THRESHOLD_FOR_SHORT_OR_CHAR) {
//private static final int NUM_SHORT_VALUES = 1 << 16;
int[] count = new int[NUM_SHORT_VALUES];
for (int i = left - 1; ++i <= right;
count[a[i] - Short.MIN_VALUE]++
);
for (int i = NUM_SHORT_VALUES, k = right + 1; k > left; ) {
while (count[--i] == 0);
short value = (short) (i + Short.MIN_VALUE);
int s = count[i];
do {
a[--k] = value;
} while (--s > 0);
}
} else { // Use Dual-Pivot Quicksort on small arrays
//进行正常的排序，和int类型的步骤一样
doSort(a, left, right, work, workBase, workLen);
}
}
char的表示范围更小，采用计数排序，和short一个道理
//private static final int NUM_CHAR_VALUES = 1 << 16;
//private static final int COUNTING_SORT_THRESHOLD_FOR_SHORT_OR_CHAR = 3200;
static void sort(char[] a, int left, int right,
char[] work, int workBase, int workLen) {
// Use counting sort on large arrays
if (right - left > COUNTING_SORT_THRESHOLD_FOR_SHORT_OR_CHAR) {
int[] count = new int[NUM_CHAR_VALUES];
for (int i = left - 1; ++i <= right;
count[a[i]]++
);
for (int i = NUM_CHAR_VALUES, k = right + 1; k > left; ) {
while (count[--i] == 0);
char value = (char) i;
int s = count[i];
do {
a[--k] = value;
} while (--s > 0);
}
} else { // Use Dual-Pivot Quicksort on small arrays
doSort(a, left, right, work, workBase, workLen);
}
}
byte sort
//private static final int COUNTING_SORT_THRESHOLD_FOR_BYTE = 29;
//private static final int NUM_BYTE_VALUES = 1 << 8;
//若待排数组长度大于29，则进行计数排序，若小于29则进行经典插入排序
static void sort(byte[] a, int left, int right) {
// Use counting sort on large arrays
if (right - left > COUNTING_SORT_THRESHOLD_FOR_BYTE) {
int[] count = new int[NUM_BYTE_VALUES];
for (int i = left - 1; ++i <= right;
count[a[i] - Byte.MIN_VALUE]++
);
for (int i = NUM_BYTE_VALUES, k = right + 1; k > left; ) {
while (count[--i] == 0);
byte value = (byte) (i + Byte.MIN_VALUE);
int s = count[i];
do {
a[--k] = value;
} while (--s > 0);
}
} else { // Use insertion sort on small arrays
for (int i = left, j = i; i < right; j = ++i) {
byte ai = a[i + 1];
while (ai < a[j]) {
a[j + 1] = a[j];
if (j-- == left) {
break;
}
}
a[j + 1] = ai;
}
}
}
float sort
/**
* 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
*/
//对folat数组进行排序时会进行一些排查
static void sort(float[] a, int left, int right,
float[] work, int workBase, int workLen) {
/*
* Phase 1: Move NaNs to the end of the array.
*/
//跳过末尾非数字
while (left <= right && Float.isNaN(a[right])) {
--right;
}
//将所有非数字移至最后
for (int k = right; --k >= left; ) {
float ak = a[k];
if (ak != ak) { // a[k] is NaN
a[k] = a[right];
a[right] = ak;
--right;
}
}
/*
* Phase 2: Sort everything except NaNs (which are already in place).
*/
//进行正常排序，步骤和整形大致相同，有一点点不一样，在后面说明
doSort(a, left, right, work, workBase, workLen);
/*
* Phase 3: Place negative zeros before positive zeros.
*/
int hi = right;
/*
* Find the first zero, or first positive, or last negative element.
*/
//找到第一个0.0f
while (left < hi) {
int middle = (left + hi) >>> 1;
float middleValue = a[middle];
if (middleValue < 0.0f) {
left = middle + 1;
} else {
hi = middle;
}
}
/*
* Skip the last negative value (if any) or all leading negative zeros.
*/
//找到第一个正0.0f
while (left <= right && Float.floatToRawIntBits(a[left]) < 0) {
++left;
}
/*
* Move negative zeros to the beginning of the sub-range.
*
* Partitioning:
*
* +----------------------------------------------------+
* | < 0.0 | -0.0 | 0.0 | ? ( >= 0.0 ) |
* +----------------------------------------------------+
* ^ ^ ^
* | | |
* left p k
*
* Invariants:
*
* all in (*, left) < 0.0
* all in [left, p) == -0.0
* all in [p, k) == 0.0
* all in [k, right] >= 0.0
*
* Pointer k is the first index of ?-part.
*/
//将所有的负0.0f移到正0.0f前面
for (int k = left, p = left - 1; ++k <= right; ) {
float ak = a[k];
if (ak != 0.0f) {
break;
}
if (Float.floatToRawIntBits(ak) < 0) { // ak is -0.0f
a[k] = 0.0f;
a[++p] = -0.0f;
}
}
}
//浮点数和其它排序不同的地方在sort(float[] a, int left, int right, boolean leftmost)方法里
private static void sort(float[] 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) {
float 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) {
float 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;
}
float 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]) { float t = a[e2]; a[e2] = a[e1]; a[e1] = t; }
if (a[e3] < a[e2]) { float 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]) { float 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]) { float 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.
*/
float pivot1 = a[e2];
float 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; ) {
float 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; ) {
float 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].
*/
//在这里,int排序如下
/*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];
}
*/
//作者的注释已经很明显了，在浮点数情况下a[less]=pivot1可能并不完
//全正确
a[less] = a[great];
++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.
*/
float 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;
}
float 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] = a[great];
}
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);
}
}
double sort和float sort基本没有区别