Ruby数组转换

Question

我有一串数字:

s = "12345678910"

正如您所看到的那样,数字1到10按递增顺序列出.我想将它转换为这些数字的数组:

a = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

我该怎么做？

Answer 1

这个怎么样:

a = ["123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475767778798081828384858687888990919293949596979899"]
b = a.first.each_char.map {|n| n.to_i }
if b.size > 8
  c = b[0..8]
  c += b[9..b.size].each_slice(2).map(&:join).map(&:to_i)
end

# It would yield as follows:
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99]

对于99以后的后续数字,请相应地修改现有谓词.