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hackbuteer1 2011-06-23
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快速排序的的快速是相对其他的排序方法来讲是快速的，你可以看下书上的比较，无论时间复杂度还是空间复杂度，可以用递归实现，也可以不用递归来实现的。。。
赵4老师 2011-06-23
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c:\Microsoft SDK\src\crt\qsort.c
/***
*qsort.c - quicksort algorithm; qsort() library function for sorting arrays
*
* Copyright (c) 1985-2001, Microsoft Corporation. All rights reserved.
*
*Purpose:
* To implement the qsort() routine for sorting arrays.
*
*******************************************************************************/
#include <cruntime.h>
#include <stdlib.h>
#include <search.h>
#include <internal.h>
/* Always compile this module for speed, not size */
#pragma optimize("t", on)
/* prototypes for local routines */
static void __cdecl shortsort(char *lo, char *hi, size_t width,
int (__cdecl *comp)(const void *, const void *));
static void __cdecl swap(char *p, char *q, size_t width);
/* this parameter defines the cutoff between using quick sort and
insertion sort for arrays; arrays with lengths shorter or equal to the
below value use insertion sort */
#define CUTOFF 8 /* testing shows that this is good value */
/***
*qsort(base, num, wid, comp) - quicksort function for sorting arrays
*
*Purpose:
* quicksort the array of elements
* side effects: sorts in place
* maximum array size is number of elements times size of elements,
* but is limited by the virtual address space of the processor
*
*Entry:
* char *base = pointer to base of array
* size_t num = number of elements in the array
* size_t width = width in bytes of each array element
* int (*comp)() = pointer to function returning analog of strcmp for
* strings, but supplied by user for comparing the array elements.
* it accepts 2 pointers to elements and returns neg if 1<2, 0 if
* 1=2, pos if 1>2.
*
*Exit:
* returns void
*
*Exceptions:
*
*******************************************************************************/
/* sort the array between lo and hi (inclusive) */
#define STKSIZ (8*sizeof(void*) - 2)
void __cdecl qsort (
void *base,
size_t num,
size_t width,
int (__cdecl *comp)(const void *, const void *)
)
{
/* Note: the number of stack entries required is no more than
1 + log2(num), so 30 is sufficient for any array */
char *lo, *hi; /* ends of sub-array currently sorting */
char *mid; /* points to middle of subarray */
char *loguy, *higuy; /* traveling pointers for partition step */
size_t size; /* size of the sub-array */
char *lostk[STKSIZ], *histk[STKSIZ];
int stkptr; /* stack for saving sub-array to be processed */
if (num < 2 || width == 0)
return; /* nothing to do */
stkptr = 0; /* initialize stack */
lo = base;
hi = (char *)base + width * (num-1); /* initialize limits */
/* this entry point is for pseudo-recursion calling: setting
lo and hi and jumping to here is like recursion, but stkptr is
preserved, locals aren't, so we preserve stuff on the stack */
recurse:
size = (hi - lo) / width + 1; /* number of el's to sort */
/* below a certain size, it is faster to use a O(n^2) sorting method */
if (size <= CUTOFF) {
shortsort(lo, hi, width, comp);
}
else {
/* First we pick a partitioning element. The efficiency of the
algorithm demands that we find one that is approximately the median
of the values, but also that we select one fast. We choose the
median of the first, middle, and last elements, to avoid bad
performance in the face of already sorted data, or data that is made
up of multiple sorted runs appended together. Testing shows that a
median-of-three algorithm provides better performance than simply
picking the middle element for the latter case. */
mid = lo + (size / 2) * width; /* find middle element */
/* Sort the first, middle, last elements into order */
if (comp(lo, mid) > 0) {
swap(lo, mid, width);
}
if (comp(lo, hi) > 0) {
swap(lo, hi, width);
}
if (comp(mid, hi) > 0) {
swap(mid, hi, width);
}
/* We now wish to partition the array into three pieces, one consisting
of elements <= partition element, one of elements equal to the
partition element, and one of elements > than it. This is done
below; comments indicate conditions established at every step. */
loguy = lo;
higuy = hi;
/* Note that higuy decreases and loguy increases on every iteration,
so loop must terminate. */
for (;;) {
/* lo <= loguy < hi, lo < higuy <= hi,
A[i] <= A[mid] for lo <= i <= loguy,
A[i] > A[mid] for higuy <= i < hi,
A[hi] >= A[mid] */
/* The doubled loop is to avoid calling comp(mid,mid), since some
existing comparison funcs don't work when passed the same
value for both pointers. */
if (mid > loguy) {
do {
loguy += width;
} while (loguy < mid && comp(loguy, mid) <= 0);
}
if (mid <= loguy) {
do {
loguy += width;
} while (loguy <= hi && comp(loguy, mid) <= 0);
}
/* lo < loguy <= hi+1, A[i] <= A[mid] for lo <= i < loguy,
either loguy > hi or A[loguy] > A[mid] */
do {
higuy -= width;
} while (higuy > mid && comp(higuy, mid) > 0);
/* lo <= higuy < hi, A[i] > A[mid] for higuy < i < hi,
either higuy == lo or A[higuy] <= A[mid] */
if (higuy < loguy)
break;
/* if loguy > hi or higuy == lo, then we would have exited, so
A[loguy] > A[mid], A[higuy] <= A[mid],
loguy <= hi, higuy > lo */
swap(loguy, higuy, width);
/* If the partition element was moved, follow it. Only need
to check for mid == higuy, since before the swap,
A[loguy] > A[mid] implies loguy != mid. */
if (mid == higuy)
mid = loguy;
/* A[loguy] <= A[mid], A[higuy] > A[mid]; so condition at top
of loop is re-established */
}
/* A[i] <= A[mid] for lo <= i < loguy,
A[i] > A[mid] for higuy < i < hi,
A[hi] >= A[mid]
higuy < loguy
implying:
higuy == loguy-1
or higuy == hi - 1, loguy == hi + 1, A[hi] == A[mid] */
/* Find adjacent elements equal to the partition element. The
doubled loop is to avoid calling comp(mid,mid), since some
existing comparison funcs don't work when passed the same value
for both pointers. */
higuy += width;
if (mid < higuy) {
do {
higuy -= width;
} while (higuy > mid && comp(higuy, mid) == 0);
}
if (mid >= higuy) {
do {
higuy -= width;
} while (higuy > lo && comp(higuy, mid) == 0);
}
/* OK, now we have the following:
higuy < loguy
lo <= higuy <= hi
A[i] <= A[mid] for lo <= i <= higuy
A[i] == A[mid] for higuy < i < loguy
A[i] > A[mid] for loguy <= i < hi
A[hi] >= A[mid] */
/* We've finished the partition, now we want to sort the subarrays
[lo, higuy] and [loguy, hi].
We do the smaller one first to minimize stack usage.
We only sort arrays of length 2 or more.*/
if ( higuy - lo >= hi - loguy ) {
if (lo < higuy) {
lostk[stkptr] = lo;
histk[stkptr] = higuy;
++stkptr;
} /* save big recursion for later */
if (loguy < hi) {
lo = loguy;
goto recurse; /* do small recursion */
}
}
else {
if (loguy < hi) {
lostk[stkptr] = loguy;
histk[stkptr] = hi;
++stkptr; /* save big recursion for later */
}
if (lo < higuy) {
hi = higuy;
goto recurse; /* do small recursion */
}
}
}
/* We have sorted the array, except for any pending sorts on the stack.
Check if there are any, and do them. */
--stkptr;
if (stkptr >= 0) {
lo = lostk[stkptr];
hi = histk[stkptr];
goto recurse; /* pop subarray from stack */
}
else
return; /* all subarrays done */
}
...
叁叁玖玖 2011-06-23
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LZ不要光看到代码的复杂度，快排是在所有数量级（O（nlogn））的排序方法中，其平衡性能最好的。
xiaohuh421 2011-06-23
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[Quote=引用楼主 udbwcso 的回复:]
内部排序
快速排序要用递归，应该是既费时又费空间的排序方法，那么为什么叫快速排序呢？
[/Quote]
楼主说它"既费时又费空间",你这是相对的.
为什么叫快排? 因为它速度快啊,书上你看下它的时间复杂度公式就知道了
内部排序
快速排序要用递归，应该是既费时又费空间的排序方法，那么为什么叫快速排序呢？
[/Quote]
楼主说它"既费时又费空间",你这是相对的.
为什么叫快排? 因为它速度快啊,书上你看下它的时间复杂度公式就知道了
luciferisnotsatan 2011-06-23
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内部排序是啥？
快排，这个是名字。快排通常的时间是 O(nlogn)，最坏是O(n^2)。而且实现比较简单，一般来说比堆排快。你可以不用递归来写
快排，这个是名字。快排通常的时间是 O(nlogn)，最坏是O(n^2)。而且实现比较简单，一般来说比堆排快。你可以不用递归来写
至善者善之敌 2011-06-23
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递归就是为了实现高效。。。。。。。。。
www_adintr_com 2011-06-23
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你也可以不用递归来实现呀
