假设我们有一个字符串，并且我们想要找到每个字母的最大重复序列。

例如，给定样本输入：

"acbaabbbaaaaacc"

样本输入的输出可以是：

a=5
c=2
b=3

规则：

• 您的代码可以是函数或程序-供您选择
• 输入可以通过标准输入，文件或函数参数
• 输出应仅包含出现在输入中的字符
• 输入最大长度为1024
• 输出顺序无关紧要，但必须以[char] = [最大重复序列] [定界符]的形式打印
• 字符串可以包含任何字符

比赛于世界标准时间23:59星期四结束。

8086机器代码，82 80

x.com文件内容：

B7 3D 89 DF B1 80 F3 AA 0D 0A 24 B4 01 CD 21 42
38 D8 74 F7 38 17 77 02 88 17 88 C3 31 D2 3C 0D
75 E9 BF 21 3D B1 5E 31 C0 F3 AE E3 EE 4F BB 04
01 8A 05 D4 0A 86 E0 0D 30 30 89 47 02 3C 30 77
04 88 67 03 43 89 3F 89 DA B4 09 CD 21 47 EB D7

它仅支持最多99个字符的重复。

源代码（用作debug.com汇编程序的输入），带注释！

a
    mov bh, 3d         ; storage of 128 bytes at address 3d00
    mov di, bx
    mov cl, 80
    rep stosb          ; zero the array
    db 0d 0a 24
; 10b
    mov ah, 1
    int 21             ; input a char
    inc dx             ; calculate the run length
    cmp al, bl         ; is it a repeated character?
    je  10b
    cmp [bx], dl       ; is the new run length greater than previous?
    ja  11a
    mov [bx], dl       ; store the new run length
; 11a
    mov bl, al         ; remember current repeating character
    xor dx, dx         ; initialize run length to 0
    cmp al, d          ; end of input?
    jne 10b            ; no - repeat
    mov di, 3d21       ; start printing run lengths with char 21
    mov cl, 5e         ; num of iterations = num of printable characters
; 127
    xor ax, ax
    repe scasb         ; look for a nonzero run length
    jcxz 11b           ; no nonzero length - exit
    dec di
    mov bx, 104        ; address of output string
    mov al, [di]       ; read the run length
    aam                ; convert to decimal
    xchg al, ah
    or  ax, 3030
    mov [bx+2], ax
    cmp al, 30         ; was it less than 10?
    ja  145
    mov [bx+3], ah     ; output only one digit
    inc bx             ; adjust for shorter string
; 145
    mov [bx], di       ; store "x=" into output string
    mov dx, bx         ; print it
    mov ah, 9
    int 21
    inc di
    jmp 127            ; repeat
; 150

rcx 50
n my.com
w
q

以下是一些我认为很有趣的高尔夫技巧：

• 数组的地址是 3d00，其中3d的ASCII码在哪里=。通过这种方式，地址为字符数组的入口x是3d78。当解释为2个字符的字符串时，为x=。
• 输出缓冲区在地址104; 它会覆盖不再需要的初始化代码。行尾序列0D 0A 24作为无害代码执行。
• aam这里的说明不提供任何打高尔夫球的信息，尽管可以...
• 将数字写两次，首先假定它大于10，然后更正它是否较小。
• 出口指令的地址晦涩难懂11b，C3运气好的话，其中包含了所需的机器代码。

CJam，27 26 25字节

l:S_&{'=L{2$+_S\#)}g,(N}/

在线尝试。

例

$ cjam maxseq.cjam <<< "acbaabbbaaaaacc"
a=5
c=2
b=3

怎么运行的

l:S       " Read one line from STDIN and store the result in “S”.                   ";
_&        " Intersect the string with itself to remove duplicate characters.        ";
{         " For each unique character “C” in “S”:                                   ";
  '=L     " Push '=' and ''.                                                        ";
  {       "                                                                         ";
    2$+_  " Append “C” and duplicate.                                               ";
    S\#)  " Get the index of the modified string in “S” and increment it.           ";
  }g      " If the result is positive, there is a match; repeat the loop.           ";
  ,       " Retrieve the length of the string.                                      ";
  (       " Decrement to obtain the highest value that did result in a match.       ";
  N       " Push a linefeed.                                                        ";
}/        "                                                                         ";

J-52个字节

好了，再次简单的方法。

f=:([,'=',m=:":@<:@#@[`(]m~[,{.@[)@.(+./@E.))"0 1~~.

说明：

f=:([,'=',m=:":@<:@#@[`(]m~[,{.@[)@.(+./@E.))"0 1~~.
                                                 ~~. Create a set of the input and apply it as the left argument to the following.
   ([,'=',m=:":@<:@#@[`(]m~[,{.@[)@.(+./@E.))"0 1    The function that does the work
                                             "0 1    Apply every element from the left argument (letters) with the whole right argument (text).
                                  @.(+./@E.)         Check if the left string is in right string.
                       (]m~[,{.@[)                   If yes, add one letter to the left string and recurse.
             ":@<:@#@[                               If not, return (length of the left string - 1), stringified.
    [,'=',                                           Append it to the letter + '='

例：

   f 'acbaabbbaaaaacc'
a=5
c=2
b=3
   f 'aaaabaa'
a=4
b=1

如果允许自由格式的输出（与其他许多答案一样），我也有一个45字节的版本。这些框代表一个框列表（是的，它们的打印方式是这样的，尽管SE的行高将它们打断了）。

   f=:([;m=:<:@#@[`(]m~[,{.@[)@.(+./@E.))"0 1~~.
   f 'acbaabbbaaaaacc'
┌─┬─┐
│a│5│
├─┼─┤
│c│2│
├─┼─┤
│b│3│
└─┴─┘
   f 'aaaabaabba'
┌─┬─┐
│a│4│
├─┼─┤
│b│2│
└─┴─┘

Ruby，72岁

(a=$*[0]).chars.uniq.map{|b|puts [b,a.scan(/#{b}+/).map(&:size).max]*?=}

这将从命令行参数中获取输入，然后输出至stdout。

GolfScript，26个字节

:s.&{61{2$=}s%1,/$-1=,n+}%

在线尝试。

说明：

• :s将输入字符串保存在变量中，s以备后用。
• .&提取输入中的唯一字符，{ }%然后循环中的其余代码对其进行迭代。
• 61 将数字61（等号的ASCII码）压入堆栈中当前字符的顶部，以用作输出定界符。
• {2$=}s%接受字符串s，如果其字符等于要迭代的当前字符，则将其字符替换为1；否则将其替换为0。（它还将当前字符留在堆栈上以进行输出。）
• 1,/ 接受这个由一和零组成的字符串，并将其拆分为零。
• $对结果子串进行排序，-1=提取最后一个子串（由于它们都由相同字符的重复组成，因此最长），然后,返回该子串的长度。
• n+ 字符串化长度，并在其后添加换行符。

附言 如果输出中的等号是可选的，则61可以省略（用2$代替1$），总长度为24个字节：

:s.&{{1$=}s%1,/$-1=,n+}%

CoffeeScript，109个字节

我喜欢正则表达式。

f=(s)->a={};a[t[0]]=t.length for t in s.match(/((.)\2*)(?!.*\1)/g).reverse();(k+'='+v for k,v of a).join '\n'

这是您可以在浏览器控制台中尝试的已编译JavaScript

f = function(s) {
  var a, t, _i, _len, _ref;
  a = {};
  _ref = s.match(/((.)\2*)(?!.*\1)/g).reverse();
  for (_i = 0, _len = _ref.length; _i < _len; _i++) {
    t = _ref[_i];
    a[t[0]] = t.length;
  }
  return a;
};

那你可以打电话

f("acbaabbbaaaaacc")

要得到

c=2
a=5
b=3

Pyth，24 25 26（或29）

=ZwFY{Z=bkW'bZ~bY)p(Yltb

测试可以在这里完成：链接

输出格式为：

('a', 5)
('c', 2)
('b', 3)

说明：

=Zw              Store one line of stdin in Z
FY{Z             For Y in set(Z):
=bk              b=''
W'bZ             while b in Z:
~bY              b+=Y
)                end while
p(Yltb           print (Y, len(b)-1)

蟒蛇：

k=""
Z=copy(input())
for Y in set(Z):
 b=copy(k)
 while (b in Z):
  b+=Y
 print(_tuple(Y,len(tail(b))))

为了获得正确的（a = 5）输出，请使用：

=ZwFY{Z=bkW'bZ~bY)p++Y"="`ltb

29个字符

C，126个 125 119字节

l,n,c[256];main(p){while(~(p=getchar()))n*=p==l,c[l=p]=c[p]>++n?c[p]:n;for(l=256;--l;)c[l]&&printf("%c=%d\n",l,c[l]);}

运行：

$ gcc seq.c 2>& /dev/null
$ echo -n 'acbaabbbaaaaacc' | ./a.out
c=2
b=3
a=5

Mathematica，74 72 69

Print[#[[1,1]],"=",Max[Tr/@(#^0)]]&/@Split@Characters@#~GatherBy~Max&

% @ "acbaabbbaaaaacc"

a=5
c=2
b=3

不是很好，但是字符串不是Mathematica的最佳区域。越来越好了。:-)

C# (LinQPad)

146

This is tsavino's answer but shorter. Here, I used Distinct() instead of GroupBy(c=>c). Also the curly braces from the foreach-loop are left out:

void v(string i){foreach(var c in i.Distinct())Console.WriteLine(c+"="+(from Match m in Regex.Matches(i,"["+c+"]+")select m.Value.Length).Max());}

136

I tried using a lambda expression instead of the normal query syntax but since I needed a Cast<Match> first, the code became 1 character longer... Anyhow, since it can be executed in LinQPad, you can use Dump() instead of Console.WriteLine():

void v(string i){foreach(var c in i.Distinct())(c+"="+(from Match m in Regex.Matches(i,"["+c+"]+")select m.Value.Length).Max()).Dump();}

Further study of the code got me thinking about the Max(). This function also accepts a Func. This way I could skip the Select part when using the lambda epxression:

void v(string i){foreach(var c in i.Distinct())(c+"="+Regex.Matches(i,"["+c+"]+").Cast<Match>().Max(m=>m.Value.Length)).Dump();}

Thus, final result:

128

Update:

Thanks to the tip from Dan Puzey, I was able to save another 6 characters:

void v(string i){i.Distinct().Select(c=>c+"="+Regex.Matches(i,"["+c+"]+").Cast<Match>().Max(m=>m‌​.Value.Length)).Dump();}

Length:

122

Python 3 (70)

s=input()
for c in set(s):
 i=1
 while c*i in s:i+=1
 print(c,'=',i-1)

Even golfed Python can be very readable. I think this code is fully idiomatic except for single-letter variables and a one-line while loop.

Example runs:

>>> helloworld
e = 1
d = 1
h = 1
l = 2
o = 1
r = 1
w = 1
>>> acbaabbbaaaaacc
a = 5
c = 2
b = 3

Ruby, 58

h={}
gets.scan(/(.)\1*/){h[$1]=[h[$1]||0,$&.size].max}
p h

Takes input from STDIN, outputs it to STDOUT in the form {"a"=>5, "c"=>2, "b"=>3}

C# in LINQPad - 159 Bytes

Well, at least I beat T-SQL ;P Won't beat anyone else, but I thought I'd share it anyway.

void v(string i){foreach(var c in i.GroupBy(c=>c)){Console.WriteLine(c.Key+"="+(from Match m in Regex.Matches(i,"["+c.Key+"]+")select m.Value.Length).Max());}}

Usage:

v("acbaabbbaaaaacc");

Suggestions are always welcome!

Powershell 80 77 72

$x=$args;[char[]]"$x"|sort -u|%{"$_="+($x-split"[^$_]"|sort)[-1].length}

You need to run it on console...

Perl - 65 71 76 characters

My first code golf!

For each answer, copy to golf.pl and run as:

echo acbaabbbaaaaacc | perl golf.pl

My shortest solution prints each character as many times as it appears, since that is not prohibited by the rules.

$_=$i=<>;for(/./g){$l=length((sort$i=~/$_*/g)[-1]);print"$_=$l
"}

My next-shortest solution (85 90 characters) only prints each character once:

<>=~s/((.)\2*)(?{$l=length$1;$h{$2}=$l if$l>$h{$2}})//rg;print"$_=$h{$_}
"for keys %h

F# - 106

let f s=
 let m=ref(Map.ofList[for c in 'a'..'z'->c,0])
 String.iter(fun c->m:=(!m).Add(c,(!m).[c]+1))s;m

In FSI, calling

f "acbaabbbaaaaacc"

gives

val it : Map<char,int> ref =
  {contents =
    map
      [('a', 8); ('b', 4); ('c', 3); ('d', 0); ('e', 0); ('f', 0); ('g', 0);
       ('h', 0); ('i', 0); ...];}

However, to print it without the extra information, call it like this:

f "acbaabbbaaaaacc" |> (!) |> Map.filter (fun _ n -> n > 0)

which gives

val it : Map<char,int> = map [('a', 8); ('b', 4); ('c', 3)]

Javascript, 116 bytes

y=x=prompt();while(y)r=RegExp(y[0]+'+','g'),alert(y[0]+'='+x.match(r).sort().reverse()[0].length),y=y.replace(r,'')

Sample output:

lollolllollollllollolllooollo
l=4
o=3

acbaabbbaaaaacc
a=5
c=2
b=3

helloworld
h=1
e=1
l=2
o=1
w=1
r=1
d=1 

T-SQL (2012) 189 171

Edit: removed ORDER BY because rules allow any output order.

Takes input from a CHAR variable, @a, and uses a recursive CTE to create a row for each character in the string and figures out sequential occurrences.

After that, it's a simple SELECT and GROUP BY with consideration for the order of the output.

Try it out on SQL Fiddle.

WITH x AS(
    SELECT @a i,''c,''d,0r,1n
    UNION ALL 
    SELECT i,SUBSTRING(i,n,1),c,IIF(d=c,r+1,1),n+1
    FROM x
    WHERE n<LEN(i)+2
)
SELECT d+'='+LTRIM(MAX(r))
FROM x
WHERE n>2
GROUP BY d

Assigning the variable:

DECLARE @a CHAR(99) = 'acbaabbbaaaaacc';

Sample output:

a=5
c=2
b=3

Haskell - 113 120 bytes

import Data.List
main=interact$show.map(\s@(c:_)->(c,length s)).sort.nubBy(\(a:_)(b:_)->a==b).reverse.sort.group

Tested with

$ printf "acbaabbbaaaaacc" | ./sl
[('a',5),('b',3),('c',2)]

JavaScript [83 bytes]

prompt().match(/(.)\1*/g).sort().reduce(function(a,b){return a[b[0]]=b.length,a},{})

^{Run this code in the browser console.}

For input "acbaabbbaaaaacc" the console should output "Object {a: 5, b: 3, c: 2}".

JavaScript - 91

for(i=0,s=(t=prompt()).match(/(.)\1*/g);c=s[i++];)t.match(c+c[0])||alert(c[0]+'='+c.length)

EDIT: My first solution obeys the rules, but it prints several times single char occurrences like abab => a=1,b=1,a=1,b=1 so I came out with this (101 chars), for those not satisfied with my first one:

for(i=0,s=(t=prompt()).match(/((.)\2*)(?!.*\1)/g);c=s[i++];)t.match(c+c[0])||alert(c[0]+'='+c.length)

Julia, 85

f(s)=(l=0;n=1;a=Dict();[c==l?n+=1:(n>get(a,l,1)&&(a[l]=n);n=1;l=c) for c in s*" "];a)
julia> f("acbaabbbaaaaacc")
{'a'=>5,'c'=>2,'b'=>3}

Python3 - 111, 126, 115 114 111 bytes

Executable code that will read 1 line (only use lowercase letters a-z)

d={}.fromkeys(map(chr,range(97,123)),0)
for c in input():d[c]+=1
[print("%s=%d"%(p,d[p]))for p in d if d[p]>0]

Edit: Excluded unnecessary output on request from @Therare

The output looks nice

~/codegolf $ python3 maxseq.py 
helloworld
l=3
o=2
h=1
e=1
d=1
w=1
r=1

JavaScript - 141 137 125

I don't like regex :)

function g(a){i=o=[],a=a.split('');for(s=1;i<a.length;){l=a[i++];if(b=l==a[i])s++;if(!b|!i){o[l]=o[l]>s?o[l]:s;s=1}}return o}

Run

console.log(g("acbaabbbaaaaacc"));

outputs

[ c: 2, a: 5, b: 3 ]

Javascript, 109 104 100 98 bytes

function c(s){q=l={};s.split('').map(function(k){q[k]=Math.max(n=k==l?n+1:1,q[l=k]|0)});return q}

Example usage:

console.log(c("aaaaaddfffabbbbdb"))

outputs:

{ a: 5, d: 2, f: 3, b: 4 }

PHP, 104 102 96

<?php function _($s){while($n=$s[$i++]){$a[$n]=max($a[$n],$n!=$s[$i-2]?$v=1:++$v);}print_r($a);}

usage

_('asdaaaadddscc');

printed

Array ( [a] => 4 [s] => 1 [d] => 3 [c] => 2 )

Java 247

import java.util.*;public class a{public static void main(String[]a){Map<Character, Integer> m = new HashMap<>();for(char c:a[0].toCharArray()){Integer v=m.get(c);m.put(c,v==null?1:v+1);}for(char c:m.keySet())System.out.println(c+"="+m.get(c));}}

C 169

Iterates each printable character in ASCII table and counts max from input string.

#define N 128
int c,i,n;
char Y[N],*p;
int main(){gets(Y);
for(c=33;c<127;c++){p=Y;n=0,i=0;while(*p){if(*p==c){i++;}else{n=(i>n)?i:n;i=0;}p++;}
if(n>0) printf("%c=%d\n",c,n);}
}

JavaScript 116

prompt(x={}).replace(/(.)\1*/g,function(m,l){n=m.length
if(!x[l]||x[l]<n)x[l]=n})
for(k in x)console.log(k+'='+x[k])

Groovy - 80 chars

Based on this clever answer by xnor :

t=args[0];t.toSet().each{i=0;
while(t.contains(it*++i));
println "${it}=${i-1}"}

Output:

$ groovy Golf.groovy abbcccdddd
d=4
b=2
c=3
a=1

Ungolfed:

t=args[0]

t.toSet().each { c ->
    i=0
    s=c

    // repeat the char c with length i
    // e.g. "b", "bb", "bbb", etc
    // stop when we find a length that is not in t:
    // this is the max + 1
    while (t.contains(s)) {
        i++
        s=c*i
    }
    println "${c}=${i-1}"
}