让面试官满意的排序算法（图文解析）
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这种排序算法能够让面试官面露微笑
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这种排序算法集各排序算法之大成
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这种排序算法逻辑性十足
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这种排序算法能够展示自己对Java底层的了解
这种排序算法出自Vladimir Yaroslavskiy、Jon Bentley和Josh Bloch三位大牛之手，它就是JDK的排序算法——java.util.DualPivotQuicksort（双支点快排）
DualPivotQuicksort
先看一副逻辑图（如有错误请大牛在评论区指正）
插排指的是改进版插排——哨兵插排
快排指的是改进版快排——双支点快排
DualPivotQuickSort没有Object数组排序的逻辑，此逻辑在Arrays中，好像是归并+Tim排序
图像应该很清楚：对于不同的数据类型，Java有不同的排序策略：
- byte、short、char 他们的取值范围有限，使用计数排序占用的空间也不过256/65536个单位，只要排序的数量不是特别少（有一个计数排序阈值，低于这个阈值的话就没有不要用空间换时间了），都应使用计数排序
- int、long、float、double 他们的取值范围非常的大，不适合使用计数排序
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float和double 他们又有特殊情况：
- NAN（not a number），NAN不等于任何数字，甚至不等于自己
- +0.0，-0.0，float和double无法精确表示十进制小数，我们所看到的十进制小数其实都是取得近似值，因而会有+0.0（接近0的正浮点数）和-0.0（接近0的负浮点数)，在排序流程中统一按0来处理，因而最后要调整一下-0.0和+0.0的位置关系
- Object
计数排序
计数排序是以空间换时间的排序算法，它时间复杂度O(n)，空间复杂度O(m)（m为排序数值可能取值的数量），只有在范围较小的时候才应该考虑计数排序
（源码以short为例）
int[] count = new int[NUM_SHORT_VALUES]; //1 << 16 = 65536，即short的可取值数量
//计数，left和right为数组要排序的范围的左界和右界
//注意，直接把
for (int i = left - 1; ++i <= right;count[a[i] - Short.MIN_VALUE]++);
//排序
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);
}
哨兵插排
当数组元素较少时，时间O(n2)和O(logn)其实相差无几，而插排的空间占用率要少于快排和归并排序，因而当数组元素较少时（<插排阈值），优先使用插排
哨兵插排是对插排的优化，原插排每次取一个值进行遍历插入，而哨兵插排则取两个，较大的一个（小端在前的排序）作为哨兵，当哨兵遍历到自己的位置时，另一个值可以直接从哨兵当前位置开始遍历，而不用再重头遍历
只画了静态图，如果有好的绘制Gif的工具请在评论区告诉我哦
我们来看一下源码：
if (leftmost) {
//传统插排（无哨兵Sentinel）
//遍历
//循环向左比较（<左侧元素——换位）-直到大于左侧元素
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;
}
//哨兵插排
} else {
//如果一开始就是排好序的——直接返回
do {
if (left >= right) {
return;
}
} while (a[++left] >= a[left - 1]);
//以两个为单位遍历，大的元素充当哨兵，以减少小的元素循环向左比较的范围
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;
}
//确保最后一个元素被排序
int last = a[right];
while (last < a[--right]) {
a[right + 1] = a[right];
}
a[right + 1] = last;
}
return;
双支点快排
重头戏：双支点快排！
快排虽然稳定性不如归并排序，但是它不用复制来复制去，省去了一段数组的空间，在数组元素较少的情况下稳定性影响也会下降（>插排阈值 ，<快排阈值），优先使用快排
双支点快排在原有的快排基础上，多加一个支点，左右共进，效率提升
看图：
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第一步，取支点
注意：如果5个节点有相等的任两个节点，说明数据不够均匀，那就要使用单节点快排
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快排
源码（int为例，这么长估计也没人看）
// Inexpensive approximation of length / 7
// 快排阈值是286 其7分之一小于等于1/8+1/64+1
int seventh = (length >> 3) + (length >> 6) + 1;
// 获取分成7份的五个中间点
int e3 = (left + right) >>> 1; // The midpoint
int e2 = e3 - seventh;
int e1 = e2 - seventh;
int e4 = e3 + seventh;
int e5 = e4 + seventh;
// 保证中间点的元素从小到大排序
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
//点彼此不相等——分三段快排，否则分两段
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];
while (a[++less] < pivot1);
while (a[--great] > pivot2);
/*
* Partitioning:
*
* left part center part right part
* +--------------------------------------------------------------+
* | < pivot1 | pivot1 <= && <= pivot2 | ? | > pivot2 |
* +--------------------------------------------------------------+
* ^ ^ ^
* | | |
* less k great
*/
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);
}
归并排序
你不会以为元素多（>快排阈值）就一定要用归并了吧？
错！元素多时确实对算法的稳定性有要求，可是如果这些元素能够稳定快排呢？
开发JDK的大牛显然考虑了这一点：他们在归并排序之前对元素进行了是否能稳定快排的判断：
- 如果数组本身几乎已经排好了（可以看出几段有序数组的拼接），那还排什么，理一理返回就行了
- 如果出现连续33个相等元素——使用快排（实话说，我没弄明白为什么，有无大牛给我指点迷津？）
//判断结构是否适合归并排序
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 {
//连续MAX_RUN_LENGTH（33）个相等元素，使用快排
for (int m = MAX_RUN_LENGTH; ++k <= right && a[k - 1] == a[k]; ) {
if (--m == 0) {
sort(a, left, right, true);
return;
}
}
}
//count达到MAX_RUN_LENGTH，使用快排
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;
}
归并排序源码
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;
}
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