希尔排序算法原理与实现

1.问题描述

输入：n个数的序列<a1,a2,a3,...,an>。

输出：原序列的一个重排<a1*,a2*,a3*,...,an*>；，使得a1*<=a2*<=a3*<=...<=an*。

2. 问题分析

例如，假设有这样一组数[ 13 14 94 33 82 25 59 94 65 23 45 27 73 25 39 10 ]，如果我们以步长为5开始进行排序，我们可以通过将这列表放在有5列的表中来更好地描述算法，这样他们就应该看起来是这样：

13 14 94 33 82
25 59 94 65 23
45 27 73 25 39
10

然后我们对每列进行排序：

10 14 73 25 23
13 27 94 33 39
25 59 94 65 82
45

将上述四行数字，依序接在一起时我们得到：[ 10 14 73 25 23 13 27 94 33 39 25 59 94 65 82 45 ].这时10已经移至正确位置了，然后再以3为步长进行排序：

排序之后变为：

最后以1步长进行排序（此时就是简单的插入排序了）。

对于每一列的排序，可以采用任意一个算法，本文采用变形的InsertSort( CVector<T> &vec,int start, int end, int step = 1 )。

3. 算法实现

template <typename T>
void InsertSort( CVector<T> &vec,int start, int end, int step = 1 )
{
    for ( size_t i=start+step; i<vec.GetSize(); i+=step )
    {
        T temp = vec[i];
        int j = i-step;

        while ( j >= start && vec[j] > temp )
        {
            vec[j+step] = vec[j];
            j-=step;
        }

        vec[j+step] = temp;
    }
}

template <typename T>
void ShellSort( CVector<T> &vec )
{
    const int gaps[10] = {1, 5, 19, 41, 109, 209, 505, 929, 2161, 3905};
    int i_gap = 9;
    int size = vec.GetSize();

    while( gaps[i_gap] >=size-1  )
        i_gap--;

    for( int gap = gaps[i_gap]; i_gap>=0; gap=gaps[--i_gap] )
    {
        //traversal each element in block
        for ( int i=0; i<gap; i++ )
        {
            //ShellSortPart( vec, i, gap  );
            InsertSort<int>( vec, i, size-1, gap );
        }
    }
}

测试：

#define  DATA_MAGNITUDE 100

double random(double start, double end)
{
    return start+(end-start)*rand()/(RAND_MAX + 1.0);
}

int main(int argc, char **argv)
{
    CVector<int> vec1(10,2);
    CVector<int> vec2(DATA_MAGNITUDE);
    CVector<char> vec_txt(1000,\'a\');

    //====================================================================
    srand( unsigned(time(0)));
    for ( int i=0; i<DATA_MAGNITUDE; i++ )
    {
        //vec2.PushBack( (int)rand()%DATA_MAGNITUDE );
        vec2.PushBack( (int)random(0, DATA_MAGNITUDE) );
    }
    unsigned int size = 20<DATA_MAGNITUDE? 20:DATA_MAGNITUDE;
    for ( size_t i =0; i < size; i++ )
    {
        cout<<vec2[i]<<" ";
    }
    cout<<endl;

    //InsertSort<int>( vec2, 0, vec2.GetSize()-1 );
    //BubbleSort<int>( vec2 );
    //SelectSort<int>( vec2 );
    ShellSort<int>( vec2 );

    for ( size_t i =0; i < size; i++ )
    {
        cout<<vec2[i]<<" ";
    }
    cout<<endl;
    return 0;
}

4. 算法分析

希尔排序的性能与选取的步长直接相关。

General term (k ≥ 1)	Concrete gaps	Worst-case time complexity	Author and year of publication
$\lfloor N / 2^k \rfloor$	$\left\lfloor\frac{N}{2}\right\rfloor, \left\lfloor\frac{N}{4}\right\rfloor, \ldots, 1$	$\Theta(N^2)$ [when N=2^p]	Shell, 1959^[1]
$2 \lfloor N / 2^{k+1} \rfloor + 1$	$2 \left\lfloor\frac{N}{4}\right\rfloor + 1, \ldots, 3, 1$	$\Theta(N^{3/2})$	Frank & Lazarus, 1960^[5]
$2^k - 1$	$1, 3, 7, 15, 31, 63, \ldots$	$\Theta(N^{3/2})$	Hibbard, 1963^[6]
$2^k + 1$ , prefixed with 1	$1, 3, 5, 9, 17, 33, 65, \ldots$	$\Theta(N^{3/2})$	Papernov & Stasevich, 1965^[7]
successive numbers of the form $2^p 3^q$	$1, 2, 3, 4, 6, 8, 9, 12, \ldots$	$\Theta(N \log^2 N)$	Pratt, 1971^[8]
$(3^k - 1) / 2$ , not greater than $\lceil N / 3 \rceil$	$1, 4, 13, 40, 121, \ldots$	$\Theta(N^{3/2})$	Knuth, 1973^[9]
$\prod \limits_{\scriptscriptstyle 0\le q<r\atop \scriptscriptstyle q\neq(r^2+r)/2-k}a_q, \hbox{where}$ $r = \left\lfloor \sqrt{2k+\sqrt{2k}} \right\rfloor,$ $a_q=\min\{n\in\mathbb{N}\colon n\ge(5/2)^{q+1},$ $\forall p\colon0\le p<q\Rightarrow\gcd(a_p,n)=1\}$	$1, 3, 7, 21, 48, 112, \ldots$	$O(N e^\sqrt{8\ln(5/2)\ln N})$	Incerpi & Sedgewick, 1985^[10]
$4^k + 3\cdot2^{k-1} + 1$ , prefixed with 1	$1, 8, 23, 77, 281, \ldots$	$O(N^{4/3})$	Sedgewick, 1986^[3]
$9(4^{k-1}-2^{k-1})+1, 4^{k+1}-6\cdot2^k+1$	$1, 5, 19, 41, 109, \ldots$	$O(N^{4/3})$	Sedgewick, 1986^[3]
$h_k = \max\left\{\left\lfloor 5h_{k-1}/11 \right\rfloor, 1\right\}, h_0 = N$	$\left\lfloor \frac{5N}{11} \right\rfloor, \left\lfloor \frac{5}{11}\left\lfloor \frac{5N}{11} \right\rfloor\right\rfloor, \ldots, 1$	?	Gonnet & Baeza-Yates, 1991^[11]
$\left\lceil \frac{9^k-4^k}{5\cdot4^{k-1}} \right\rceil$	$1, 4, 9, 20, 46, 103, \ldots$	?	Tokuda, 1992^[12]
unknown	$1, 4, 10, 23, 57, 132, 301, 701$	?	Ciura, 2001^[13]