Question

我正在格式化电话号码，短划线后不能有一位数字，即123-555-5555或12-34，但不能是123-4。答案中也可以包含任何字母字符。

Here是我的答案。

class CodeTestException < Exception; end

# driver method
def phone_format(s)
  string = s.to_s.gsub(/[^0-9]/, '') # force input to string
  string_length = string.length

  # ensure that the return type stays consistent and make sure that there isn't
  # one digit by itself as specified by the API
  return string unless string_length > 2

  format_phone_string(string, string_length)
end

private

def format_phone_string(string, string_length)
  formatted_string = ""
  early_dash = string_length % 3 == 1
  skip_dash = false

  # start index at 1 for easier comprehension
  i = 0

  string.each_char do |char|
    formatted_string << char
    break if i == string_length

    i += 1

    formatted_string << '-' && skip_dash = true if early_dash && (i == string_length - 2)

    formatted_string << '-' if i % 3 == 0 && !skip_dash
  end

  formatted_string
end

raise CodeTestException unless phone_format("(+1) 888 33x19") == "188-833-19"
raise CodeTestException unless phone_format("555 123 1234") == "555-123-12-34"
raise CodeTestException unless phone_format("(+1)") == '1'
raise CodeTestException unless phone_format(nil) == ""
raise CodeTestException unless phone_format("") == ""

有人告诉我答案不是递归或扫描。我相信O(n)是我可以度过的最佳时间。我发现其他一些解决方案是线性时间的较高倍数。谁能击败我的解决方案？

格式化字符串的脚本速度测试

0 个答案: