具有O(n*log(n))时间和O(1)空间复杂性的稳定比较排序

Question

在浏览维基百科的排序算法列表时，我注意到没有一种稳定的比较排序具有O(n*log(n))（最坏情况）时间复杂性和O(1)（最坏条件）空间复杂性。这看起来确实是一个理论边界，但我找不到更多关于它的信息

如何证明这一点

注意：我知道比较排序的O(n*log(n))最坏情况时间复杂度的下限。

Answer 1

不管那篇文章怎么说，就地稳定的Merge Sort可以变成O(n log n)。

这篇论文解释了实现它的两种方法。

Answer 2

这里是一个完整的C++实现，我从这里的C实现转换而来：

#include <algorithm>
#include <functional>
template<class It>
It rotate(It begin, It const middle, It end)
{
    typename std::iterator_traits<It>::difference_type i = 0, j;
    if (begin != middle && middle != end)
    {
        while ((i = std::distance(begin, middle)) !=
               (j = std::distance(middle, end)))
        {
            It k = middle;
            std::advance(
                k,
                std::max(
                    typename std::iterator_traits<It>::difference_type(),
                    j - i));
            std::swap_ranges(k, end, begin);
            if (i > j) { std::advance(begin, j); }
            else { std::advance(end, -i); }
        }
    }
    return std::swap_ranges(middle - i, middle, middle);
}
template<class It, class Less>
It bsearch(
    It begin, It left, It right,
    typename std::iterator_traits<It>::difference_type n,
    Less &less)
{
    while (left < right)
    {
        It const middle = left + std::distance(left, right) / 2;
        bool const b = !less(
            *(begin + (std::distance(middle, begin) + n)),
            *middle);
        (b ? left : right) = middle + b;
    }
    return left;
}
template<class It, class Less>
void merge(It const begin, It const middle, It const end, Less &less)
{
    bool naive_insertion_optimization = false;
    if (naive_insertion_optimization && std::distance(begin, end) < 0)
    {
        for (It i = middle; i != end; ++i)
        {
            for (It p = i; p != begin; --p)
            {
                if (!less(*p, *(p - 1)))
                {
                    break;
                }
                using std::iter_swap;
                iter_swap(p, p - 1);
            }
        }
    }
    else if (begin < middle && middle < end)
    {
        typename std::iterator_traits<It>::difference_type const
            half = std::distance(begin, end) / 2,
            left = std::distance(begin, middle),
            right = std::distance(middle, end);
        It const midpoint = begin + half;
        bool const b = left > right;
        It const i = bsearch(
            begin,
            b ? midpoint - right : begin,
            b ? midpoint : middle,
            half + left - 1,
            less);
        rotate(i, middle, begin + (std::distance(i, middle) + half));
        merge(begin, i, midpoint, less);
        merge(midpoint, midpoint + std::distance(i, middle), end, less);
    }
}
template<class It, class Less>
void sort(It const begin, It const end, Less &less)
{
    if (std::distance(begin, end) > 1)
    {
        It const middle = begin + std::distance(begin, end) / 2;
        sort(begin, middle, less);
        sort(middle, end, less);
        merge(begin, middle, end, less);
    }
}
template<class It>
void sort(It const begin, It const end)
{
    std::less<typename std::iterator_traits<It>::value_type> less;
    return sort(begin, end, less);
}