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原文出处： 陶邦仁
二分法查找算法，算法思想：当数据量很大适宜采用该方法。采用二分法查找时，数据需是排好序的。 基本思想：假设数据是按升序排序的，对于给定值x，从序列的中间位置开始比较，如果当前位置值等于x，则查找成功；若x小于当前位置值，则在数列的前半段中查找；若x大于当前位置值则在数列的后半段中继续查找，直到找到为止。
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//针对int类型数组的二分法查找，key为要查找数的下标 private static int binarySearch0(int[] a, int fromIndex, int toIndex, int key) { int low = fromIndex; int high = toIndex - 1; while (low <= high) { int mid = (low + high) >>> 1;//无符号左移一位，相当于除以二 int midVal = a[mid]; if (midVal < key) low = mid + 1; else if (midVal > key) high = mid - 1; else return mid; // key found } return -(low + 1); // key not found. } |
针对引用类型数组采取的算法是归并排序，算法思想：归并（Merge）排序法是将两个（或两个以上）有序表合并成一个新的有序表，即把待排序序列分为若干个子序列，每个子序列是有序的。然后再把有序子序列合并为整体有序序列。
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private static final int INSERTIONSORT_THRESHOLD = 7;//插入排序门槛 public static void sort(Object[] a) { Object[] aux = (Object[])a.clone(); mergeSort(aux, a, 0, a.length, 0); } //归并排序 private static void mergeSort(Object[] src, Object[] dest, int low, int high, int off) { int length = high - low; if (length < INSERTIONSORT_THRESHOLD) { //若数组长度小于7，则用冒泡排序 for (int i=low; i<high; i++) for (int j=i; j>low && ((Comparable) dest[j-1]).compareTo(dest[j])>0; j--) swap(dest, j, j-1); return; } // Recursively sort halves of dest into src int destLow = low; int destHigh = high; low += off; high += off; int mid = (low + high) >>> 1; //无符号左移一位， mergeSort(dest, src, low, mid, -off); mergeSort(dest, src, mid, high, -off); // If list is already sorted, just copy from src to dest. This is an // optimization that results in faster sorts for nearly ordered lists. if (((Comparable)src[mid-1]).compareTo(src[mid]) <= 0) { System.arraycopy(src, low, dest, destLow, length); return; } // Merge sorted halves (now in src) into dest for(int i = destLow, p = low, q = mid; i < destHigh; i++) { if (q >= high || p < mid && ((Comparable)src[p]).compareTo(src[q])<=0) dest[i] = src[p++]; else dest[i] = src[q++]; } } |
采取的是快速排序算法，算法思想：通过一趟排序将要排序的数据分割成独立的两部分，其中一部分的所有数据都比另外一部分的所有数据都要小，然后再按此方法对这两部分数据分别进行快速排序，整个排序过程可以递归进行，以此达到整个数据变成有序序列。
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/** * Swaps x[a] with x[b]. */ private static void swap(int x[], int a, int b) { int t = x[a]; x[a] = x[b]; x[b] = t; } public static void sort(int[] a) { sort1(a, 0, a.length); } private static int med3(int x[], int a, int b, int c) {//找出三个中的中间值 return (x[a] < x[b] ? (x[b] < x[c] ? b : x[a] < x[c] ? c : a) : (x[b] > x[c] ? b : x[a] > x[c] ? c : a)); } /** * Sorts the specified sub-array of integers into ascending order. */ private static void sort1(int x[], int off, int len) { // Insertion sort on smallest arrays if (len < 7) {//采用冒泡排序 for (int i=off; i<len+off; i++) for (int j=i; j>off && x[j-1]>x[j]; j--) swap(x, j, j-1); return; } //采用快速排序 // Choose a partition element, v int m = off + (len >> 1); // Small arrays, middle element if (len > 7) { int l = off; int n = off + len - 1; if (len > 40) { // Big arrays, pseudomedian of 9 int s = len/8; l = med3(x, l, l+s, l+2*s); m = med3(x, m-s, m, m+s); n = med3(x, n-2*s, n-s, n); } m = med3(x, l, m, n); // Mid-size, med of 3 } int v = x[m]; // Establish Invariant: v* (<v)* (>v)* v* int a = off, b = a, c = off + len - 1, d = c; while(true) { while (b <= c && x[b] <= v) { if (x[b] == v) swap(x, a++, b); b++; } while (c >= b && x[c] >= v) { if (x[c] == v) swap(x, c, d--); c--; } if (b > c) break; swap(x, b++, c--); } // Swap partition elements back to middle int s, n = off + len; s = Math.min(a-off, b-a ); vecswap(x, off, b-s, s); s = Math.min(d-c, n-d-1); vecswap(x, b, n-s, s); // Recursively sort non-partition-elements if ((s = b-a) > 1) sort1(x, off, s); if ((s = d-c) > 1) sort1(x, n-s, s); } |
针对double，float类型数组排序，采取了先把所有的数组元素值为-0.0d的转换成0.0d，再利用快速排序排好序，最后再还原。
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public static void sort(double[] a) { sort2(a, 0, a.length); } private static void sort2(double a[], int fromIndex, int toIndex) { //static long doubleToLongBits(double value) //根据 IEEE 754 浮点双精度格式 ("double format") 位布局，返回指定浮点值的表示形式。 final long NEG_ZERO_BITS = Double.doubleToLongBits(-0.0d); /* * The sort is done in three phases to avoid the expense of using * NaN and -0.0 aware comparisons during the main sort. */ /* * Preprocessing phase: Move any NaN‘s to end of array, count the * number of -0.0‘s, and turn them into 0.0‘s. */ int numNegZeros = 0; int i = fromIndex, n = toIndex; while(i < n) { if (a[i] != a[i]) { //这段搞不懂，源代码怪怪的，感觉多此一举 double swap = a[i]; a[i] = a[--n]; a[n] = swap; } else { if (a[i]==0 && Double.doubleToLongBits(a[i])==NEG_ZERO_BITS) { a[i] = 0.0d; numNegZeros++; } i++; } } // Main sort phase: quicksort everything but the NaN‘s sort1(a, fromIndex, n-fromIndex); // Postprocessing phase: change 0.0‘s to -0.0‘s as required if (numNegZeros != 0) { int j = binarySearch0(a, fromIndex, n, 0.0d); // posn of ANY zero do { j--; } while (j>=0 && a[j]==0.0d); // j is now one less than the index of the FIRST zero for (int k=0; k<numNegZeros; k++) a[++j] = -0.0d; } } |
http://www.importnew.com/19952.html#comment-499513
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原文地址：http://www.cnblogs.com/softidea/p/5614037.html
