本文简述了 计算 MaxSortedDistance 的一些方法

所谓 MaxSortedDistance,是指一个无序数组经过排序后相邻元素的最大差值.

举个例子,如果给定无序数组:

1000 , 13 , 1 , 7 , 5 , 3 , 100 1000, 13, 1, 7, 5, 3, 100 1000,13,1,7,5,3,100

那么该数组的 MaxSortedDistance 即为 900,因为该数组经过排序后为:

1 , 3 , 5 , 7 , 13 , 100 , 1000 1, 3, 5, 7, 13, 100, 1000 1,3,5,7,13,100,1000

其中相邻元素最大的差值为:

1000 − 100 = 900 1000 - 100 = 900 1000−100=900

最简单直白的实现方法自然是直接给数组排序,然后计算最大的相邻元素差值即可:

function get_max_sorted_distance_sort(num_table)
    table.sort(num_table)
    
    local max_sort_distance = 0
    
    for i = 2, #num_table do
        if num_table[i] - num_table[i - 1] > max_sort_distance then
            max_sort_distance = num_table[i] - num_table[i - 1]
        end
    end
    
    return max_sort_distance
end

代码简单直白,通俗易懂.

但是排序数组的时间复杂度比较高,有可能进一步降低吗?

一种方式便是借鉴 计数排序 的算法思路来给数组"排序":

function get_max_sorted_distance_count(num_table)
    local num = #num_table
    if num > 1 then
        local min = num_table[1]
        local max = min
        
        for i = 2, num do
            local val = num_table[i]
            if val < min then
                min = val
            elseif val > max then
                max = val
            end
        end
        
        local num_buffer = {}
        
        for i = 1, max - min + 1 do
            num_buffer[i] = 0
        end
        
        for i = 1, num do
            -- no need to '+= 1' here(just need '=1')
            num_buffer[num_table[i] - min + 1] = 1
        end
        
        local max_sort_distance = 0
        local cur_sort_distance = 0
        
        for i = 2, max - min + 1 do
            if num_buffer[i - 1] == 0 and num_buffer[i] ~= 0 then
                if cur_sort_distance + 1 > max_sort_distance then
                    max_sort_distance = cur_sort_distance + 1
                end
                cur_sort_distance = 0
            elseif num_buffer[i - 1] == 0 and num_buffer[i] == 0 then
                cur_sort_distance = cur_sort_distance + 1
            elseif num_buffer[i - 1] ~= 0 and num_buffer[i] == 0 then
                cur_sort_distance = 1
            elseif num_buffer[i - 1] ~= 0 and num_buffer[i] ~= 0 then
                -- no need to set here
                --cur_sort_distance = 0
            end
        end
    
        return max_sort_distance
    end
    
    return 0
end

但是计数排序的算法思路至少有两个问题:

1. 数组元素相差较大时,空间消耗较大
2. 仅适用于整形类型数组

对于此,我们可以采用桶排序的算法思路来解决:

function get_max_sorted_distance_bucket(num_table)
    local num = #num_table
    if num > 1 then
        local min = num_table[1]
        local max = min
        
        for i = 2, num do
            local val = num_table[i]
            if val < min then
                min = val
            elseif val > max then
                max = val
            end
        end
        
        local bucket_buffer = {}
        local bucket_range = (max - min) / (num - 1)
        
        -- check nearly zero here
        local epsilon = 1e-5
        if bucket_range <= epsilon then
            return 0
        end
        
        for i = 1, num do
            bucket_buffer[i] = {}
        end
        
        for i = 1, num do
            local index_raw = (num_table[i] - min) / bucket_range
            local index = math.floor(index_raw + 0.5) + 1
            if bucket_buffer[index] then
                if bucket_buffer[index].min then
                    if num_table[i] < bucket_buffer[index].min then
                        bucket_buffer[index].min = num_table[i]
                    end
                else
                    bucket_buffer[index].min = num_table[i]
                end
                
                if bucket_buffer[index].max then
                    if num_table[i] > bucket_buffer[index].max then
                        bucket_buffer[index].max = num_table[i]
                    end
                else
                    bucket_buffer[index].max = num_table[i]
                end
            else
                print("[get_max_sort_distance_bucket]out of bucket size error occur ...")
            end
        end
        
        local max_sort_distance = 0
        local pre_bucket_index = 0
        
        for i = 1, num do
            -- seems no need to check min and max in same bucket
            --[[
            if bucket_buffer[i].min and bucket_buffer[i].max then
                local bucket_distance = bucket_buffer[i].max - bucket_buffer[i].min
                if bucket_distance > max_sort_distance then
                    max_sort_distance = bucket_distance
                end
            end
            --]]
            
            if pre_bucket_index > 0 then
                if bucket_buffer[pre_bucket_index].max and bucket_buffer[i].min then
                    local bucket_distance = bucket_buffer[i].min - bucket_buffer[pre_bucket_index].max
                    if bucket_distance > max_sort_distance then
                        max_sort_distance = bucket_distance
                    end
                end
            end
            
            if bucket_buffer[i].max then
                pre_bucket_index = i
            end
        end
    
        return max_sort_distance
    end
    
    return 0
end

测试代码:

local t = { 1000, 13, 1, 7, 5, 3, 100 }

print(get_max_sorted_distance_count(t))
print(get_max_sorted_distance_bucket(t))
print(get_max_sorted_distance_sort(t))

很久没有动手写博文了,原因能"编"出来很多,大抵都是些老生长谈的东西,不赘述了,今天也仅是兴盛所致写了一篇,希望以后能多写一些了,是为记.