Question

我必须使用名为Barometer Operation的方法找到Merge Sort的指令计数。

什么是晴雨表操作？

选择“晴雨表指令”
计数=执行气压计指令的次数。
搜索算法：气压计指令（x == L [j]？）。
排序算法：气压计指令（L [i]＆lt; = L [j]？）。

例如： -

     for(int i=0;i < n;i++)
       A[i] = i + 1

循环体

气压计操作= +

count（+）= n

所以现在Merge Sort的问题在于它是递归算法，我不知道如何选择一条特定指令，这样我就可以计算执行特定指令的次数。

Answer 1

感谢 @aphex 和 @ SGM1 向我展示正确的方向。我想解决问题的方法如下：

我使用了与 aphex 相同的代码。但只有解决方案的问题是 aphex 忘记在最坏的情况下保持它。所以根据我的气压计操作/说明必须

first(left) <= first(right)

因为这是进行实际比较的地方。晴雨表操作员将是“＆lt; =”，其中进行实际比较。并且这个运算符通过两个方法递归调用（n / 2）次，这两个方法将列表分开然后主合并操作恰好发生（n-1）次。因此在求解求和方程后它给出（nlog（n）+ c）

来自Wiki的代码，很少通过 aphex

进行修改

function merge_sort(list m, var count)
    if(length(m))>0
        count++;

// if list size is 0 (empty) or 1, consider it sorted and return it
// (using less than or equal prevents infinite recursion for a zero length m)
if length(m) <= 1
    return m

// else list size is > 1, so split the list into two sublists
var list left, right
var integer middle = length(m) / 2
for each x in m before middle
     add x to left
for each x in m after or equal middle
     add x to right
// recursively call merge_sort() to further split each sublist
// until sublist size is 1
left = merge_sort(left,count)
right = merge_sort(right,count)
// merge the sublists returned from prior calls to merge_sort()
// and return the resulting merged sublist
return merge(left, right)

function merge(left, right)
var list result
while length(left) > 0 or length(right) > 0
    if length(left) > 0 and length(right) > 0
        if first(left) <= first(right) // My Barometer Operation
            append first(left) to result
            left = rest(left)
        else
            append first(right) to result
            right = rest(right)
    else if length(left) > 0
        append first(left) to result
        left = rest(left)
    else if length(right) > 0
        append first(right) to result
        right = rest(right)
end while
return result

Answer 2

让我们使用Wikipedia中的代码。你需要指望分裂部分。分割是在merge_sort(list m)中进行的。您需要将计数器添加为输入参数（var count），如下所示。该计数器必须为count=0。每次调用方法和列表lenght>0时，都会将计数器增加1.

function merge_sort(list m, var count)
      // Count here
    if(length(m))>0
            count++;

    // if list size is 0 (empty) or 1, consider it sorted and return it
    // (using less than or equal prevents infinite recursion for a zero length m)
    if length(m) <= 1
        return m

    // else list size is > 1, so split the list into two sublists
    var list left, right
    var integer middle = length(m) / 2
    for each x in m before middle
         add x to left
    for each x in m after or equal middle
         add x to right
    // recursively call merge_sort() to further split each sublist
    // until sublist size is 1
    left = merge_sort(left,count)
    right = merge_sort(right,count)
    // merge the sublists returned from prior calls to merge_sort()
    // and return the resulting merged sublist
    return merge(left, right)

function merge(left, right)
    var list result
    while length(left) > 0 or length(right) > 0
        if length(left) > 0 and length(right) > 0
            if first(left) <= first(right)
                append first(left) to result
                left = rest(left)
            else
                append first(right) to result
                right = rest(right)
        else if length(left) > 0
            append first(left) to result
            left = rest(left)
        else if length(right) > 0
            append first(right) to result
            right = rest(right)
    end while
    return result

对于维基百科示例，您需要调用伪代码，如merge_sort([38,27,43,3,9,82,10], 0)。

递归算法中的气压计操作

2 个答案: