Arrays排序算法
2021/7/11 11:06:44
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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 if (run[count] > left && a[run[count]] >= a[run[count] - 1]) { count--; } /* * The array is not highly structured, * use Quicksort instead of merge sort. */ if (++count == MAX_RUN_COUNT) { sort(a, left, right, true); return; } } // These invariants should hold true: // run[0] = 0 // run[<last>] = right + 1; (terminator) if (count == 0) { // A single equal run return; } else if (count == 1 && run[count] > right) { // Either a single ascending or a transformed descending run. // Always check that a final run is a proper terminator, otherwise // we have an unterminated trailing run, to handle downstream. return; } right++; if (run[count] < right) { // Corner case: the final run is not a terminator. This may happen // if a final run is an equals run, or there is a single-element run // at the end. Fix up by adding a proper terminator at the end. // Note that we terminate with (right + 1), incremented earlier. run[++count] = right; } // 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; int o = ao; ao = bo; bo = o; } }
LeetCode
力扣题库:75. 颜色分类、力扣官方题解
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