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这种排序算法能够让面试官面露微笑
这种排序算法集各排序算法之大成
这种排序算法逻辑性十足
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)
想看以往学习内容的朋友
可以看我的GitHub: https://github.com/Meng997998/AndroidJX
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
float和double他们又有特殊情况:
Object
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); }
当数组元素较少时,时间O(n 2)和O(log n)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;
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:
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
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+1int seventh = (length >> 3) + (length >> 6) + 1;// 获取分成7份的五个中间点int e3 = (left + right) >>> 1; // The midpointint 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; } } } }// Pointersint less = left; // The index of the first element of center partint 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); }
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:
//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 sortedfor (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 mergingint[] b; // temp array; alternates with aint ao, bo; // array offsets from 'left'int blen = right - left; // space needed for bif (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; }// Mergingfor (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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