Question

我决定尝试在Lua中重新创建一个排序算法，只是为了好玩。它适用于某些值，但是在具有重复数字的较大值中，它不能正常工作。我想知道为什么会这样。

这里是代码（我添加了评论以便于查看）：

numbers = "9876554321" -- Number to sort, I will refer to this as the "string to sort"
length = tonumber(string.len(numbers)) -- The length of the string to sort
changes = 1 -- The next few lines are just variables
totalchanges = 0
iterations = 0
newNumbers = numbers
print("")
while changes ~= 0 do -- This repeats until the number of changes equals zero
  iterations = iterations + 1 -- This variable doesn't do anything for the code, it's only used at the end
  changes = 0
  print(newNumbers)
  x = 1 -- More variables
  y = 0
  z = 0
  w = false
  while x < length do -- While the x variable is less than the length of the string to sort, repeat this code
    y = tonumber(string.sub(newNumbers, x, x)) -- The 'x'th place in the string
    z = tonumber(string.sub(newNumbers, x + 1, x + 1)) -- The 'x + 1'th place in the string
    w = y > z -- If the 'x'th place is larger than the 'x + 1'th place, this value is set to true
    print(y) -- Print statements
    print(z)
    print(w)
    if w == true then -- Checks if y > z is true
      newNumbers = string.gsub(newNumbers, tostring(z), tostring(y), 1)  -- The next two lines swap around the numbers if y > z
      newNumbers = string.gsub(newNumbers, tostring(y), tostring(z), 1)
      changes = changes + 1 -- Increases the number of changes
    end
    print("")
    x = x + 1 -- Increases the x value, this lets the code know where to check in the string
  end
  totalchanges = totalchanges + changes -- This variable is like the iterations variable, it's only used for the last print statements
end
print("") -- These last print statements are just showing information about the sort
print("Sorting complete!")
print("Original string: " .. tostring(numbers))
print("Sorted string: " .. tostring(newNumbers))
print("Total changes made: " .. tostring(totalchanges))
print("Total iterations used: " .. tostring(iterations))

有谁知道问题是什么？

Answer 1

newNumbers = string.gsub(newNumbers, tostring(z), tostring(y), 1)      
newNumbers = string.gsub(newNumbers, tostring(y), tostring(z), 1)

这两行在很多层面都是错误的。

我们只从其中一个开始：假设z == 5，并在您的示例代码中有5个，每个 {{序列中的1}}将变为5，而不仅仅是您尝试交换的那个。

接下来是一个更明显的问题，即您首先将所有y转换为z s，然后将所有y转换为y s。这里的问题是你也变成z你刚刚变成z的{{1}}，所以基本上

和不

即使情况并非如此，也不要将字符串用于那种东西。 Lua中的字符串与通常的字符串有点不同¹。将字符串转换成序列，用它做一些事情，最后将它连接成一个字符串，你会好得多。

另请注意，您可以直接比较两个字符，Lua会按字典顺序对它们进行比较，因此无需aaaayzaa -> aaaayyaa -> aaaazzaa所有内容。 aaaayzaa -> aaaazyaa就像tostring()一样。

¹ Lua Performance Tips by Roberto Ierusalimschy, page 22 "About Strings"

Lua程序不起作用

1 个答案: