JAVA源码走读（二）二分查找与Arrays类
给数组赋值：通过fill方法。
对数组排序：通过sort方法,按升序。
比较数组：通过equals方法比较数组中元素值是否相等。
查找数组元素：通过binarySearch方法能对排序好的数组进行二分查找法操作。
使用如下：
int[] array = new int[5]; //填充数组 Arrays.fill(array, 5); System.out.println("填充数组：Arrays.fill(array, 5)："); test.output(array); //将数组的第2和第3个元素赋值为8 Arrays.fill(array, 2, 4, 8); System.out.println("将数组的第2和第3个元素赋值为8：Arrays.fill(array, 2, 4, 8)："); test.output(array); int[] array1 = {7,8,3,2,12,6,3,5,4}; //对数组的第2个到第6个进行排序进行排序 Arrays.sort(array1,2,7); System.out.println("对数组的第2个到第6个元素进行排序进行排序：Arrays.sort(array,2,7)："); test.output(array1); //对整个数组进行排序 Arrays.sort(array1); System.out.println("对整个数组进行排序：Arrays.sort(array1)："); test.output(array1); //比较数组元素是否相等 System.out.println("比较数组元素是否相等:Arrays.equals(array, array1):"+"\n"+Arrays.equals(array, array1)); int[] array2 = array1.clone(); System.out.println("克隆后数组元素是否相等:Arrays.equals(array1, array2):"+"\n"+Arrays.equals(array1, array2)); //使用二分搜索算法查找指定元素所在的下标（必须是排序好的，否则结果不正确） Arrays.sort(array1); System.out.println("元素3在array1中的位置：Arrays.binarySearch(array1, 3)："+"\n"+Arrays.binarySearch(array1, 3)); //如果不存在就返回负数 System.out.println("元素9在array1中的位置：Arrays.binarySearch(array1, 9)："+"\n"+Arrays.binarySearch(array1, 9)); int aaa[] = {5,4,7,3,8,1}; sort(aaa,0,aaa.length - 1); for(int i = 0 ;i<aaa.length;i++){ System.out.println(aaa[i]); }
源码解析：package test;
import java.util.Arrays; public class test { private static final int QUICKSORT_THRESHOLD = 286; private static final int MAX_RUN_COUNT = 67; private static final int MAX_RUN_LENGTH = 33; private static final int INSERTION_SORT_THRESHOLD = 47; public static void main(String[] args){ int aaa[] = {5,4,7,3,8,1}; sort(aaa,0,aaa.length - 1); for(int i = 0 ;i<aaa.length;i++){ System.out.println(aaa[i]); } } public static void sort(int[] a, int left, int right) { // 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) { 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. */ if (++count == MAX_RUN_COUNT) { sort(a, left, right, true); return; } } // Check special cases if (run[count] == right++) { // The last run contains one element run[++count] = right; } else if (count == 1) { // The array is already sorted return; } /* * Create temporary array, which is used for merging. * Implementation note: variable "right" is increased by 1. */ int[] b; byte odd = 0; for (int n = 1; (n <<= 1) < count; odd ^= 1); if (odd == 0) { b = a; a = new int[b.length]; for (int i = left - 1; ++i < right; a[i] = b[i]); } else { b = new int[a.length]; } // 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] <= a[q]) { b[i] = a[p++]; } else { b[i] = a[q++]; } } run[++last] = hi; } if ((count & 1) != 0) { for (int i = right, lo = run[count - 1]; --i >= lo; b[i] = a[i] ); run[++last] = right; } int[] t = a; a = b; b = t; } } /** * 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 */ private static void sort(int[] a, int left, int right, boolean leftmost) { int length = right - left + 1; // 数组后一个值与前一个值进行循环比较，通过j--进行循环替换，例如：初始值为 5 4 7 3 8 1 ， 4 5 7 3 8 1 ， 4 5 3 7 8 1 ， 4 3 5 7 8 1 if (length < INSERTION_SORT_THRESHOLD) { if (leftmost) { /* * Traditional (without sentinel) insertion sort, * optimized for server VM, is used in case of * the leftmost part. */ 插入排序
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); } } }
今日太晚，明日再干~
