与大多数语言不同，Python a<b<c会像在数学中一样进行评估，实际上是比较三个数字，而不是将布尔值a<b与进行比较c。用C（以及许多其他语言）编写此代码的正确方法是a<b && b<c。

在这个挑战中，您的任务是将这样任意长度的比较链从Python /直观表示扩展到如何用其他语言编写。

技术指标

• 您的程序必须要处理运算符：==, !=, <, >, <=, >=。
• 输入将具有仅使用整数的比较链。
• 不用担心一路比较的真实性，这纯粹是语法/语法上的挑战。
• 输入将没有任何空格，以防止答案因分割空格而使解析变得琐碎。
• 但是，您的输出可能只有一个空格，要么仅包含&&，要么包含比较运算符和&&，或者两者都不包含，但是要保持一致。

测试用例

Input                  Output
---------------------------------------------------------------

3<4<5                  3<4 && 4<5
3<4<5<6<7<8<9          3<4 && 4<5 && 5<6 && 6<7 && 7<8 && 8<9
3<5==6<19              3<5 && 5==6 && 6<19
10>=5<7!=20            10>=5 && 5<7 && 7!=20
15==15==15==15==15     15==15 && 15==15 && 15==15 && 15==15

这是 代码高尔夫球，因此以字节为单位的最短代码胜出！

视网膜，33 22 17字节

-5字节感谢@MartinEnder

-2`\D(\d+)
$&&&$1

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外壳，16 14字节

在每个运算符周围打印一个空格。

wJ"&&"m←C2X3ġ±

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说明：

                  Implicit input, e.g            "3<4<5<6"
            ġ±    Group digits                   ["3","<","4","<","5","<","6"]
          X3      Sublists of length 3           [["3","<","4"],["<","4","<"],["4","<","5"],["<","5","<"],["5","<","6"]]
        C2        Cut into lists of length 2     [[["3","<","4"],["<","4","<"]],[["4","<","5"],["<","5","<"]],[["5","<","6"]]]
      m←          Take the first element of each [["3","<","4"],["4","<","5"],["5","<","6"]]
 J"&&"            Join with "&&"                 ["3","<","4","&&","4","<","5","&&","5","<","6"]
w                 Print, separates by spaces

视网膜，42 47 22字节

凯文·克鲁伊森（Kevin Cruijssen）大力打高尔夫球

(\D(\d+))(?=\D)
$1&&$2

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V，37字节

òͨ䫄0-9& ]«©¨ä«©¨„0-9& ]«©/±² ¦¦ ²³

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十六进制转储：

00000000: f2cd a8e4 ab84 302d 3926 205d aba9 a8e4  ......0-9& ]....
00000010: aba9 a884 302d 3926 205d aba9 2fb1 b220  ....0-9& ]../.. 
00000020: a6a6 20b2 b3                             .. ..

Clojure，88字节

更新：subs代替clojure.string/join。

#(subs(apply str(for[p(partition 3 2(re-seq #"(?:\d+|[^\d]+)" %))](apply str" && "p)))4)

J，59 46字节

4(}.;)_2{.\3' && '&;\]</.~0+/\@,2~:/\e.&'!<=>'

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它过去如何运作

                        (0 , }. ~:&(e.&'!<=>') }:)

我们正在寻找运营商的界限。“斩首”和“缩减”的字符串分别变为零和1（其中0是数字），然后异或在一起。前面加零以匹配长度。

                   +/\                      </. ]     Split on boundaries. 
              ┌──┬──┬─┬─┬─┬──┬──┐
              │10│>=│5│<│7│!=│20│
              └──┴──┴─┴─┴─┴──┴──┘
         3' && '&;\          Add && to infixes of 3.
              ┌────┬──┬──┬──┐
              │ && │10│>=│5 │
              ├────┼──┼──┼──┤
              │ && │>=│5 │< │
              ├────┼──┼──┼──┤
              │ && │5 │< │7 │
              ├────┼──┼──┼──┤
              │ && │< │7 │!=│
              ├────┼──┼──┼──┤
              │ && │7 │!=│20│
              └────┴──┴──┴──┘
    _2{.\                    Take even numbered rows.
;}.,                         Concatenate after dropping the first box.

Python 2中，60个 59 58 57字节

lambda s:re.sub(r'(\D(\d+))(?=\D)',r'\1&&\2',s)
import re

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木炭，29字节

≔ ηＦθ«¿›ι9«Ｆ›η!⁺&&η≔ωη»≔⁺ηιηι
≔ ηＦθ«Ｆ∧›ι9›η!⁺&&η≔⎇›ι9ω⁺ηιηι

相同基本算法的两种形式略有不同。输入字符串按字符进行迭代。找到数字后，它们将收集在变量中。当找到数字和运算符之间的边界时，将显示一个额外的“ &&”和变量，并清除该变量。变量最初被初始化为空格，以便第一个边界不会触发额外的“ &&”。

果冻，16字节

e€ØDŒg⁸ṁṡ3m2j⁾&&

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说明：

e€ØDŒg⁸ṁṡ3m2j⁾&& Uses Jelly stringification, thus full program only
eۯD             For each each char, 1 if it's in '0123456789', otherwise 0
    Œg           Split into longest runs of equal elements
      ⁸ṁ         Reshape original input like the list we just made
                 Reshaping will separate numbers from operators
        ṡ3       Get contiguous sublists of length 3
          m2     Keep every second item, starting from the first
                 This will keep every (number operator number) and
                 remove every (operator number operator) substring
            j⁾&& Join with '&&'

Java 8，46字节

s->s.replaceAll("(\\D(\\d+))(?=\\D)","$1&&$2")

说明：

在这里尝试。

s->                       // Method with String as both parameter and return-type
   s.replaceAll("(\\D(\\d+))(?=\\D)",
                "$1&&$2") //  Replace the match of the regex
                          //  with the second String
                          // End of method (implicit / single-line return-statement)

正则表达式说明：

(\D(\d+))(?=\D)   # 1) For all matches of this regex
   (\d+)          #  Capture group 2: a number (one or more digits)
(\D \d+ )         #  Capture group 1: anything but a number + the number
                  #   (`\D` will match the `=`, `!`, `<`, or `>`)
         (?=\D)   #  Look-ahead so everything after the match remains as is

 $1&&$2           # 2) Replace it with
 $1               #  Result of capture group 1 (trailing part of the equation + the number)
   &&             #  Literal "&&"
     $2           #  Result of capture group 2 (the number)

替换的分步示例：

Initial:                             10>=5<7!=20
 Match of first replacement:            =5
 Replace with:                          =5&&5
After first replacement:             10>=5&&5<7!=20
 Match of second replacement:                <7
 Replace with:                               <7&&7
After second replacement (result):   10>=5&&5<7&&7!=20

Perl 5，47 +1（-p）= 48字节

$\.=$1." && "x(!!$')while s/(\d+\D+(\d+))/$2/}{

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Ruby，37个字节

->s{s.gsub(/\D(\d+)(?=\D)/,'\&&&\1')}

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JavaScript（SpiderMonkey），43字节

f=s=>s.replace(/(\W+(\d+))(?!$)/g,"$1&&$2")

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