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这是一个家庭作业。我最终会将这段代码翻译成 MIPS 程序集,但这对我来说是容易的部分。我一直在调试这段代码好几个小时,并且一直在我的教授办公时间,但我仍然无法让我的快速排序算法工作。这是代码以及我对我认为问题区域所在位置的一些评论:

// This struct is in my .h file
typedef struct {
    // v0 points to the first element in the array equal to the pivot
    int *v0;

    // v1 points to the first element in the array greater than the pivot (one past the end of the pivot sub-array)
    int *v1;
} PartRet;

void swap(int *a, int *b) {
    int temp = *a;
    *a = *b;
    *b = temp;
}

PartRet partition(int *lo, int *hi) {
    // Will later be translating this to MIPS where 2 values can be returned. I am using a PartRet struct to simulate this.
    PartRet retval;

    // We must use the last item as the pivot
    int pivot = *hi;

    int *left = lo;

    // Take the last value before the pivot
    int *right = hi - 1;

    while (left < right) {
        while((left < hi) && (*left <= pivot)) {
            ++left;
        }

        while((right > lo) && (*right > pivot)) {
            --right;
        }

        if (left < right) {
            swap(left++, right--);
        }
    }

    // Is this correct? left will always be >= right after the while loop
    if (*hi < *left) {
        swap(left, hi);
    }

    // MADE CHANGE HERE
    int *v0 = hi;
    int *v1;

    // Starting at the left pointer, find the beginning of the sub-array where the elements are equal to the pivot
    // MADE CHANGE HERE
    while (v0 > lo && *(v0 - 1) >= pivot) {
        --v0;
    }

    v1 = v0;

    // Starting at the beginning of the sub-array where the elements are equal to the pivot, find the element after the end of this array.
    while (v1 < hi && *v1 == pivot) {
    ++v1;
    }

    if (v1 <= v0) {
        v1 = hi + 1;
    }

    // Simulating returning two values
    retval.v0 = v0;
    retval.v1 = v1;

    return retval;
}

void quicksort(int *array, int length) {
    if (length < 2) {
        return;
    }

    PartRet part = partition(array, array + length - 1);

    // I *think* this first call is correct, but I'm not sure.
    int firstHalfLength = (int)(part.v0 - array);
    quicksort(array, firstHalfLength);

    int *onePastEnd = array + length;
    int secondHalfLength = (int)(onePastEnd - part.v1);

    // I have a feeling that this isn't correct
    quicksort(part.v1, secondHalfLength);
}

我什至尝试使用在线代码示例重写代码,但要求是使用 lo 和 hi 指针,但我发现没有代码示例使用它。当我调试代码时,我最终只使代码适用于某些数组,而不是其他数组,尤其是当枢轴是数组中的最小元素时。

4

2 回答 2

1

分区代码有问题。这里有 4 个简单的测试输出,来自下面的 SSCCE 代码:

array1:
Array (Before):
[6]: 23 9 37 4 2 12
Array (First half partition):
[3]: 2 9 4
Array (First half partition):
[1]: 2
Array (Second half partition):
[1]: 9
Array (Second half partition):
[2]: 23 37
Array (First half partition):
[0]:
Array (Second half partition):
[1]: 23
Array (After):
[6]: 2 4 9 12 37 23


array2:
Array (Before):
[3]: 23 9 37
Array (First half partition):
[1]: 23
Array (Second half partition):
[1]: 9
Array (After):
[3]: 23 37 9


array3:
Array (Before):
[2]: 23 9
Array (First half partition):
[0]:
Array (Second half partition):
[1]: 23
Array (After):
[2]: 9 23


array4:
Array (Before):
[2]: 9 24
Array (First half partition):
[0]:
Array (Second half partition):
[1]: 9
Array (After):
[2]: 24 9

SSCCE 代码

#include <stdio.h>

typedef struct
{
    int *v0;    // v0 points to the first element in the array equal to the pivot
    int *v1;    // v1 points to the first element in the array greater than the pivot (one past the end of the pivot sub-array)
} PartRet;

static void dump_array(FILE *fp, const char *tag, int *array, int size)
{
    fprintf(fp, "Array (%s):\n", tag);
    fprintf(fp, "[%d]:", size);
    for (int i = 0; i < size; i++)
        fprintf(fp, " %d", array[i]);
    putchar('\n');
}

static void swap(int *a, int *b)
{
    int temp = *a;
    *a = *b;
    *b = temp;
}

static PartRet partition(int *lo, int *hi)
{
    // Will later be translating this to MIPS where 2 values can be
    // returned.  I am using a PartRet struct to simulate this.
    PartRet retval;

    // This code probably won't ever be hit as the base case in the QS
    // function will return first
    if ((hi - lo) < 1)
    {
        retval.v0 = lo;
        retval.v1 = lo + (hi - lo) - 1;
        return retval;
    }

    // We must use the last item as the pivot
    int pivot = *hi;

    int *left = lo;

    // Take the last value before the pivot
    int *right = hi - 1;

    while (left < right)
    {
        if (*left <= pivot)
        {
            ++left;
            continue;
        }

        if (*right >= pivot)
        {
            --right;
            continue;
        }

        swap(left, right);
    }

    // Is this correct? left will always be >= right after the while loop
    swap(left, hi);

    int *v0 = left;
    int *v1;

    // Starting at the left pointer, find the beginning of the sub-array
    // where the elements are equal to the pivot
    while (v0 > lo && *(v0 - 1) == pivot)
    {
        --v0;
    }

    v1 = v0;

    // Starting at the beginning of the sub-array where the elements are
    // equal to the pivot, find the element after the end of this array.
    while (v1 < hi && *v1 == pivot)
    {
        ++v1;
    }

    // Simulating returning two values
    retval.v0 = v0;
    retval.v1 = v1;

    return retval;
}

static void quicksort(int *array, int length)
{
    if (length < 2)
    {
        return;
    }

    PartRet part = partition(array, array + length - 1);

    // I *think* this first call is correct, but I'm not sure.
    int firstHalfLength = (int)(part.v0 - array);
    dump_array(stdout, "First half partition", array, firstHalfLength);
    quicksort(array, firstHalfLength);

    int *onePastEnd = array + length;
    int secondHalfLength = (int)(onePastEnd - part.v1);

    // I have a feeling that this isn't correct
    dump_array(stdout, "Second half partition", part.v1, secondHalfLength);
    quicksort(part.v1, secondHalfLength);
}

static void mini_test(FILE *fp, const char *name, int *array, int size)
{
    putc('\n', fp);
    fprintf(fp, "%s:\n", name);
    dump_array(fp, "Before", array, size);
    quicksort(array, size);
    dump_array(fp, "After", array, size);
    putc('\n', fp);
}

int main(void)
{
    int array1[] = { 23, 9, 37, 4, 2, 12 };
    enum { NUM_ARRAY1 = sizeof(array1) / sizeof(array1[0]) };
    mini_test(stdout, "array1", array1, NUM_ARRAY1);

    int array2[] = { 23, 9, 37, };
    enum { NUM_ARRAY2 = sizeof(array2) / sizeof(array2[0]) };
    mini_test(stdout, "array2", array2, NUM_ARRAY2);

    int array3[] = { 23, 9, };
    enum { NUM_ARRAY3 = sizeof(array3) / sizeof(array3[0]) };
    mini_test(stdout, "array3", array3, NUM_ARRAY3);

    int array4[] = { 9, 24, };
    enum { NUM_ARRAY4 = sizeof(array4) / sizeof(array4[0]) };
    mini_test(stdout, "array4", array4, NUM_ARRAY4);

    return(0);
}

我没有对排序代码进行任何算法更改。我只是简单地添加了dump_array()函数,并对其进行了战略调用,然后添加了mini_test()函数和main(). 这些固定装置非常有用。请注意,当输入数组的大小为 2 且顺序错误时,分区是正确的,但当大小为 2 且顺序正确时,分区已颠倒了数组元素的位置。这是有问题的!解决它,您可能已经修复了大部分其余部分。在所有 6 个排列(3 个不同的值)中使用一些 3 个元素数组;考虑使用 3 个元素和只有 2 个不同的值,以及 3 个元素和只有 1 个值。

于 2013-03-27T05:26:28.823 回答
1

注意:我发布这个只是为了让你看看不同的方法。我已经在上面的评论中指出了您的算法中的至少一个缺陷。考虑一下。

具有集成分区的传统就地快速排序通常如下所示,尽管每个人似乎都有自己的最爱。为了简单起见,我更喜欢这样(分区和递归在同一个过程中)。最重要的是,汇编的翻译非常简单,但你现在可能已经知道了:

void quicksort(int *lo, int *hi)
{
    /* early exit on trivial slice */
    size_t len = (hi - lo) + 1;
    if (len <= 1)
        return;

    /* use hi-point for storage */
    swap(lo + len/2, hi);

    /* move everything in range below pivot */
    int *pvt=lo, *left=lo;
    for(; left != hi; ++left)
        if (*left <= *hi)
            swap(left, pvt++);

    /* this is the proper spot for the pivot */
    swap(pvt, hi);

    /* recurse sublists. do NOT include pivot slot. */
    quicksort(lo, pvt-1);
    quicksort(pvt+1, hi);
}

该算法中唯一真正的变量是如何计算提取初始枢轴值的“点”。今天的许多实现都使用随机点,因为它有助于分散快速排序对近似排序列表的退化性能:

    swap(lo + (rand() % len), hi);

其他实现就像从切片中间抓取元素一样,就像我在上面的算法中所做的那样:

    swap(lo + len/2, hi);

您可能会发现有趣的是,有些人根本不抓住中间元素并进行交换,只是使用高位中的任何内容作为枢轴值。这仅将代码减少了一行,但有一个巨大的警告:它将在几乎排序或完全排序的列表上保证糟糕的交换性能(一切都会交换)。

无论如何,如果没有别的,我希望它可以帮助您了解算法的分区部分正在尝试做什么:将枢轴值以下的所有内容推到列表头部的插槽中,将其上方的所有内容推到列表的尾部,然后递归到这些分区中,但最重要的是,不要在一切片的递归中包含枢轴槽。它已经在它需要的地方(事实上,在那个时候,它是唯一可以保证的地方)。

于 2013-03-27T06:15:41.680 回答