排序算法在JDK中的应用（二）快速排序

/** * Sorts the specified range of the array by Dual-Pivot Quicksort. * 通过双轴快速排序对指定范围内的数据进行排序 * @param a the array to be sorted 被排序的数组 * @param left the index of the first element, inclusive, to be sorted 需要排序的第一个元素的位置（包括在内） * @param right the index of the last element, inclusive, to be sorted 需要排序的最后一个元素的位置（包括在内） * @param leftmost indicates if this part is the leftmost in the range leftmost表示该部分是否是范围内最左的部分 */ private static void sort(int[] a, int left, int right, boolean leftmost) { int length = right - left + 1; // Use insertion sort on tiny arrays //当数组的长度很小时就是用插入排序，INSERTION_SORT_THRESHOLD=47 if (length < INSERTION_SORT_THRESHOLD) { if (leftmost) { /* * Traditional (without sentinel) insertion sort, 传统的插入排序，不使用哨兵元素 * optimized for server VM, is used in case of 针对最左边的部分的情况进行了服务器虚拟机的优化 * 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. 跳过最长的升序情况，提高算法效率 */ do { if (left >= right) { return; } } while (a[++left] >= a[left - 1]); /* * Every element from adjoining part plays the role 在这种排序方法中相邻的每个元素都起到了哨兵的作用 * of sentinel, therefore this allows us to avoid the 这种办法可以避免我们每次迭代时都要进行左范围检查。 * left range check on each iteration. Moreover, we use 而且我们还使用了一个效率更好的算法，我们称之为“双插入排序”， * the more optimized algorithm, so called pair insertion 在快速排序的上下文中（即满足进入sort()方法的数组）他比传统的 * sort, which is faster (in the context of Quicksort) 插入排序更快 * 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; } //从这里开始是对待排序的元素进行分组处理 // Inexpensive approximation of length / 7 // 使用length/7作为近似的加权长度 int seventh = (length >> 3) + (length >> 6) + 1; /* * Sort five evenly spaced elements around (and including) the 在范围内的中心元素附近找到5个均匀间隔的元素 * center element in the range. These elements will be used for 这些元素将用于下面代码中的枢轴选择 * pivot selection as described below. The choice for spacing 根据经验，这些元素的间距能够很好的应对和处理各种各样的输入（待排序的数组） * 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 使用插入排序对这些元素进行排序 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; } } } }//分组完成 // Pointers 指针 int less = left; // The index of the first element of center part 中心部分第一个元素的位置 int great = right; // The index before the first element of right part 右边第一个元素之前的位置 //五个分位点的值各不相同 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. 使用五个分位点中的第二个和第四个作为枢轴 * These values are inexpensive approximations of the first and 因为有上面的排序 所以在这里pivot1 <= pivot2 * 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 将要排序的第一个和最后一个元素换到枢轴的位置 * locations formerly occupied by the pivots. When partitioning 当分区操作完成后，枢轴元素将和这个元素交换回到原来的位置 * is complete, the pivots are swapped back into their final 并排除到后续排序之外 * positions, and excluded from subsequent sorting. */ a[e2] = a[left]; a[e4] = a[right]; /* * Skip elements, which are less or greater than pivot values. * 筛选那些比枢轴元素更大或者更小的元素 以此来确定less和great的位置 */ 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. * 以下的forless-1开始向右遍历至great，把小于pivot1的元素移动到less左边，大于pivot2的元素移动到great右边。 */ //outer标签 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. * 在这里分开写的原因是因为前者效率更佳 * 想要了解的可以看这里的讨论：https://www.oschina.net/question/3037675_2206753 */ a[less] = ak; ++less; } else if (ak > pivot2) { // Move a[k] to right part while (a[great] > pivot2) { if (great-- == k) { break outer; } } //通过上面已知great<pivot2，但是并不知道great和pivot1的大小关系， //如果它比pivot1还小，需要移动到到less左边，否则只需要交换到k处。 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 // 将less-1的元素放到对头，great+1的元素放在队尾 //然后将pivot1放在less-1,pivot2放在great+1 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 //递归左右部分进行排序 包括已知的轴心 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. * 如果中心部分太大(大小超过了整个数组的4/7),就将内部的枢轴值交换到端点 */ if (less < e1 && e5 < great) { /* * Skip elements, which are equal to pivot values. * 调整less和great指针的位置，即跳过相等的元素 */ 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 *这下面的排序和上面的区别在于边界条件的判断 * 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]; /* * 浮点零的相关问题：https://stackoverflow.com/questions/13544342/why-do-floating-point-numbers-have-signed-zeros * Even though a[great] equals to pivot1, the 如果a[great]和pivot1是不同符号的浮点零， * assignment a[less] = pivot1 may be incorrect, 即使a[great]这个元素等于pivot1，“a[less] = pivot1”这个赋值操作也可能是不正确的 * if a[great] and pivot1 are floating-point zeros （这里是对单双精度元素排序的一个特例处理，使-0排在+0之前） * of different signs. Therefore in float and 因此在单双精度的排序算法中我们必须使用更加精确的赋值即a[less]=a[great] * 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 // 对中心部分进行递归排序 sort(a, less, great, false); } else { // Partitioning with one pivot 采用单轴分区 区别于上面那种情况 /* * Use the third of the five sorted elements as pivot. 使用5个排序好的元素中的第三个作为枢轴元素 * This value is inexpensive approximation of the median. 这个值是数组的中值近似值 */ int pivot = a[e3]; /* * Partitioning degenerates to the traditional 3-way 分区方式退化为传统的3路形式 * (or "Dutch National Flag") schema:（或者“荷兰国旗”模式） * 荷兰国旗问题（Dutch National Flag Problem）:https://www.cnblogs.com/freelancy/p/7940803.html * 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 * 以下算法与上面算法思路差不多 不再累述 * 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.对左右部分进行递归排序 * All elements from center part are equal 中间的元素都相等，所以已经排序 * and, therefore, already sorted. */ sort(a, left, less - 1, leftmost); sort(a, great + 1, right, false); } }