java源码分析:Arrays.sort

时间:2022-05-20 07:53:47
 仔细分析java的Arrays.sort(version 1.71, 04/21/06)后发现,java对primitive(int,float等原型数据)数组采用快速排序,对Object对象数组采用归并排序。
  对这一区别,sun在<<The Java Tutorial>>中做出的解释是:
  The sort operation uses a slightly optimized merge sort algorithm that is fast and stable:
   * Fast: It is guaranteed to run in n log(n) time and runs substantially faster on nearly sorted lists. Empirical tests showed it to be as fast as a highly optimized quicksort. A quicksort is generally considered to be faster than a merge sort but isn't stable and doesn't guarantee n log(n) performance.
  * Stable: It doesn't reorder equal elements. This is important if you sort the same list repeatedly on different attributes. If a user of a mail program sorts the inbox by mailing date and then sorts it by sender, the user naturally expects that the now-contiguous list of messages from a given sender will (still) be sorted by mailing date. This is guaranteed only if the second sort was stable.
  也就是说,优化的归并排序既快速(nlog(n))又稳定。
  对于对象的排序,稳定性很重要。比如成绩单,一开始可能是按人员的学号顺序排好了的,现在让我们用成绩排,那么你应该保证,本来张三在李四前面,即使他们成绩相同,张三不能跑到李四的后面去。
  而快速排序是不稳定的,而且最坏情况下的时间复杂度是O(n^2)。
  另外,对象数组中保存的只是对象的引用,这样多次移位并不会造成额外的开销,但是,对象数组对比较次数一般比较敏感,有可能对象的比较比单纯数的比较开销大很多。归并排序在这方面比快速排序做得更好,这也是选择它作为对象排序的一个重要原因之一。
  排序优化:实现中快排和归并都采用递归方式,而在递归的底层,也就是待排序的数组长度小于7时, 直接使用冒泡排序,而不再递归下去。
  分析:长度为6的数组冒泡排序总比较次数最多也就1+2+3+4+5+6=21次,最好情况下只有6次比较。而快排或归并涉及到递归调用等的开销,其时间效率在n较小时劣势就凸显了,因此这里采用了冒泡排序,这也是对快速排序极重要的优化。
  /*快速排序*/
  private static void sort1(int x[], int off, int len) {
  // Insertion sort on smallest arrays
    if (len < 7) {
      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;
    }
  // 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);//取后三个参数中的中间值
  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);
  }
  /*归并排序*/
  private static void mergeSort(Object[] src,
  Object[] dest,
  int low,
  int high,
  int off) {
  int length = high - low;
  // Insertion sort on smallest arrays
  if (length < INSERTIONSORT_THRESHOLD) {
  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;
  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++];
  }
  }