给定仅包含字母数字ASCII字符并以字母开头的输入字符串，请将每个字母运行与后面的数字运行交换。
一个运行是连续的字母或数字序列。注意，在输入字符串以字母结尾的情况下，该行程保持不变。
演练示例
例如，给定输入字符串uV5Pt3I0：
uV 5 Pt 3 I 0(uV 5) (Pt 3) (I 0)(5 uV) (3 Pt) (0 I)5uV3Pt0I例子
uV5Pt3I0 -> 5uV3Pt0I
J0i0m8 -> 0J0i8m
abc256 -> 256abc
Hennebont56Fr -> 56HennebontFr
Em5sA55Ve777Rien -> 5Em55sA777VeRien
nOoP -> nOoP
这是代码高尔夫球，因此最短的答案以字节为单位。鼓励解释。
Answers:
~ṠŒg⁸ṁṭ2/
~ṠŒg⁸ṁṭ2/ Main link. Argument: s (string)
~ Apply bitwise NOT.
Bitwise operators attempt to cast to int, so if c is a digit, this
yields ~int(c), a negative number.
If c cannot be cast to int, ~ will yield 0.
Ṡ Take the sign.
We've now mapped digits to -1, non-digits to 0.
Œg Group consecutive equal elements.
⁸ṁ Mold s as the result, grouping run of digits and runs of non-digits.
2/ Reduce all pairs of runs by...
ṭ tack, appending the first run of the pair to the second run.
感谢@Laikoni削减了2个字节
f""=""
f s|(a,(b,y))<-span(<':')<$>span(>'9')s=b++a++f y取消高尔夫：
f "" = ""
f string | (letters, afterLetters) <- span (> '9') string
, (numbers, afterNumbers) <- span (< ':') afterLetters
= numbers ++ letters ++ f afterNumbers(a,(b,y):_)<-lex<$>span(>'9')s：在线尝试！
lex运作，因此暂时不要加入。无论如何，很高兴知道序曲中有类似的内容
q/(\d+/ ò mw c q
注意：这是一个不稳定的Alpha，因此，如果此链接中断，则可以在v1.4.4中使用稍长的版本： 在线测试！
q/(\d+/ ò mw c q : Implicit input "uV5Pt3I0"
q : Split input on
/(\d+/ : runs of digits, keeping each run. (This compiles to the regex /(\d+)/g)
: This gives ["uV","5","Pt","3","I","0",""]
ò : Take every pair of items. [["uV","5"],["Pt","3"],["I","0"],[""]]
m : Map each pair by
w : reversing. [["5","uV"],["3","Pt"],["0","I"],[""]]
c : Flatten into one array. ["5","uV","3","Pt","0","I",""]
q : Join into a single string. "5uV3Pt0I"
: Implicit: output result of last expression
ò。
q{i_64>X\:X^{])[}&c}/]]2/Wf%
CJam没有正则表达式，也没有“分成数字和字母”之类的东西，所以这有点痛苦。
q e# Read the input.
{ e# Do the following for every char c:
i e# Get c's codepoint.
64> e# Check if it's greater than 64 (i.e. if it's a letter), pushing 1 or 0.
X e# Push X (variable predefined to 1).
\:X e# Store whether c was a letter or digit into X.
^{ e# If (old X) XOR (new X) is 1:
] e# Close the current array.
) e# Pull out its last character.
[ e# Open a new array.
}& e# (end if)
c e# Turn the codepoint back into a character. This also shoves it into the new array,
e# in case one was opened.
}/ e# (end for)
] e# Close the final array, since it hasn't been closed yet.
] e# Wrap the whole stack into an array.
2/ e# Split elements into groups of 2.
Wf% e# Reverse each group.
e# Implicitly flatten and print.
<L><D>=$2$1
样品运行：
bash-4.4$ gema '<L><D>=$2$1' <<< 'Em5sA55Ve777Rien'
5Em55sA777VeRien
s/([^0-9]+)([0-9]+)/\2\1/g
用-r运行。
使用捕获组并以相反的顺序替换它们。
[A-Za-z]为[^0-9]。但是，您必须将标志算作代码的一部分。
sed <command>和之间的差异sed -r <command>，因此为三个字节。
sed -f filename和sed -rf filename（或之间sed -e 'command'和sed -re 'command'）：一个字节。
s/([a-z]+)([0-9]+)|([0-9]+)([a-z]+)/\2\1\4\3/gi48个字节。否则，大致相同。
e€ØDŒg⁸ṁs2Ṛ€
说明：
e€ØDŒg⁸ṁs2Ṛ€ Accepts a string
eۯD Check if each char is in ['0'..'9']
Œg Split runs of 0s and 1s (respectively letter and digit runs)
⁸ṁ Replace with input, keeping the split
s2 Get pairs of runs, last left alone if letter run
Ṛ€ Swap each pair
for(;a&$c=$argn[$i++];$p=$c)$c<A?print$c:$s=($p<A?!print$s:$s).$c;echo$s;作为管道运行-nR或在线测试。
分解
for(;a&$c=$argn[$i++]; # loop through input
$p=$c) # 2. remember character
$c<A # 1. if digit
?print$c # then print it
:$s=($p<A # else if previous character was digit
?!print$s # then print and reset string
:$s # else do nothing
).$c; # append current character to string
echo$s; # print remaining string~的，而不是a&
s=>System.Text.RegularExpressions.Regex.Replace(s,@"(\D+)(\d+)","$2$1")在C＃中，很遗憾的正则表达式很长。
完整/格式化版本：
using System;
class P
{
static void Main()
{
Func<string, string> f = s => System.Text.RegularExpressions.Regex.Replace(s, @"(\D+)(\d+)", "$2$1");
Console.WriteLine(f("uV5Pt3I0"));
Console.WriteLine(f("J0i0m8"));
Console.WriteLine(f("abc256"));
Console.WriteLine(f("Hennebont56Fr"));
Console.WriteLine(f("Em5sA55Ve777Rien"));
Console.WriteLine(f("nOoP"));
Console.ReadLine();
}
}哦，正则表达式真的很方便...无论如何（TIO）：
#(apply str(flatten(map reverse(partition-all 2(partition-by(fn[i](< 47(int i)58))%)))))
partition-by根据该函数的返回值拆分为连续的运行，partition-all拆分为2的分区（我们将交换的对），map reverse反转它们，flatten摆脱嵌套列表结构，最后输出一个字符串。如果partition用代替partition-all而我们的块数为奇数，则最后一个将被丢弃。
原使用冗长，但乐趣(juxt second first)和(set"0123456789")，而不是reverse和ASCII整数范围。
#(apply str(flatten(map(juxt second first)(partition-all 2(partition-by(comp not(set"0123456789"))%)))))
aR-C+XL.C+XD{c.b}
将输入作为命令行参数。在线尝试！
它使用标准的正则表达式替换策略，有些打高尔夫球。
正则表达式为-C+XL.C+XD，其计算结果为`(?i)([a-z]+)(\d+)`：
XL Preset regex variable for lowercase letter: `[a-z]`
+ Apply + to the regex: `[a-z]+`
C Wrap the regex in a capturing group: `([a-z]+)`
- Apply the case-insensitive flag: `(?i)([a-z]+)`
XD Preset regex variable for digit: `\d`
+ Apply + to the regex: `\d+`
C Wrap the regex in a capturing group: `(\d+)`
. Concatenate the two regexes: `(?i)([a-z]+)(\d+)`
替代方法是{c.b}，一个回调函数将第二组（c）和第一组（b）连接起来。（函数的第一个参数a包含整个匹配项。）
这比天真短3个字节aR`(\D+)(\d+)``\2\1`。
,[>>----[---->+<<<-[>]>]>>[.[[-]<]<<[-]+>>]<[[-<<<+>>>]<<<<[-<[<]>[.[-]>]]>[-<+>]>],]<[-]<[<]>[.>]
该程序维护一个尚未输出的字母队列，并在适当时输出它们。
该程序的关键是>>----[---->+<<<-[>]>]。输入单元格右边的三个单元格从零开始。如果输入是1到63（含1和63）之间的代码点，则这会将指针向右移动一个空格，并将输入移到这个新位置的两个空格的右边。否则，指针将向右移动两个空格，新位置向右一个空格的单元格变为63，并从输入单元格中减去相同的63。这样可以将输入整齐地分为字母（65-122）和数字（48-57）。
,[ Take first input byte and start main loop
>> Move two cells to the right
----[---->+<<<-[>]>] (See above)
>> Move two cells to the right
This cell contains the input if it was a digit, and 0 if input was a letter
[ If input was a digit:
. Output digit immediately
[[-]<] Zero out digit and working cell
<<[-]+>> Set flag so we know later that we've output a digit
]
< Move one cell left
This cell contains 63 if input was a letter, and 0 if input was a digit
[ If input was a letter:
[-<<<+>>>] Add 63 back to input letter
<<<< Move to flag
[ If a digit has been output since the last letter read:
- Clear flag
<[<]> Move to start of queue
[.[-]>] Output and clear all queued letters
]
>[-<+>]> Move input to end of queue
]
,] Repeat until no input remains
<[-] Clear flag if present
<[<]> Move to start of queue
[.>] Output all queued letters
(n=NumberString;l=Length;s=Riffle[a=StringCases[#,n],b=StringSplit[#,n]];If[l@a==0,s=#,If[l@a<l@b,AppendTo[s,b[[-2;;]]]]];""<>s)&
(a,(b,y))<-span(<':')<$>span(>'9')s。