Tantangan

Diberikan dua string yang hanya berisi huruf kecil dan tanpa spasi, hasilnya harus berupa string yang lebih pendek, diikuti oleh garis bawah, diikuti oleh string yang lebih panjang dengan instance karakter pertama yang dihapus untuk setiap karakter yang dikandungnya dalam string yang lebih pendek.

Anda dapat mengasumsikan senarnya akan selalu berbeda panjang.

Kasus uji:

sale
salinewater
Result: sale_inwater (only the first 'a' in salinewater is removed)

jostling
juggle
Result: juggle_ostin (juggle is the shorter string)

juggle
juggler
Result: juggle_r (a 'g' is removed for every 'g' in the shorter string)

Aturan

Ini kode-golf, jadi jawaban tersingkat dalam byte menang!

Pyth, 13 byte

++hAlDQ\_.-HG

Cobalah online.

JavaScript (ES6), 78 75 69 Bytes

const 
     g=(x,y)=>x[y.length]?g(y,x):[...x].map(c=>y=y.replace(c,''))&&x+'_'+y
;

console.log(g.toString().length + 2);   // 69
console.log(g('sale', 'salinewater'))   // sale_inwater
console.log(g('juggle', 'juggler'))     // juggle_r
console.log(g('jostling','juggle'))     // juggle_ostin

Kerusakan

x[y.length]?g(y,x):        \\ Make sure that x is the shorter string
[...x]                     \\ Spread string in array of characters
.map(c=>y=y.replace(c,'')) \\ For each character remove its first occurence in y
&&x+'_'+y                  \\ Concat x and changed y 

Haskell, 56 55 byte

import Data.List
x%y|(0<$y)<(0<$x)=y%x|z<-y\\x=x++'_':z

-1 byte terima kasih kepada @xnor

Java 7, 262 byte

import java.util.*;String c(String z,String y){int i=0,l=y.length();if(z.length()>l)return c(y,z);List x=new ArrayList();for(;i<l;x.add(y.toCharArray()[i++]));for(Object q:z.toCharArray())x.remove(q);String r="";for(i=0;i<x.size();r+=x.get(i++));return z+"_"+r;}

Mungkin bisa bermain golf lagi hanya dengan menggunakan array tanpa daftar ..

Kasus yang tidak disatukan & uji:

Coba di sini.

import java.util.*;
class M{
  static String c(String z, String y){
    int i = 0,
        l = y.length();
    if(z.length() > l){
      return c(y, z);
    }
    List x = new ArrayList();
    for(; i < l; x.add(y.toCharArray()[i++]));
    for(Object q : z.toCharArray()){
      x.remove(q);
    }
    String r = "";
    for(i = 0; i < (x.size()); r += x.get(i++));
    return z+"_"+r;
  }

  public static void main(String[] a){
    System.out.println(c("sale", "salinewater"));
    System.out.println(c("jostling", "juggle"));
    System.out.println(c("juggle", "juggler"));
  }
}

Keluaran:

sale_inwater
juggle_ostin
juggle_r

Java 8, 156 Bytes

String a(String x,String y){int l=x.length(),m=y.length();String b=l>m?x:y,s=m<l?y:x;for(char c:s.toCharArray()){b=b.replaceFirst(""+c,"");}return s+"_"+b;}

Ini mungkin bisa golf lagi.

Kasus yang tidak disatukan & uji

interface A {
    static String a(String x,String y){
        int l=x.length(),m=y.length();
        String b=l>m?x:y,s=m<l?y:x;
        for(char c:s.toCharArray()){
            b=b.replaceFirst(""+c,"");
        }
        return s+"_"+b;
    }

    static void main(String[]a) {
        System.out.println(a("sale","salinewater"));
        System.out.println(a("jostling","juggle"));
        System.out.println(a("juggle","juggler"));
    }
}

Ruby, 65 byte

->a,b{a,b=b,a if a.size>b.size;a.chars.map{|e|b.sub!e,""};a+?_+b}

ungolfed

->a, b{
  a, b = b, a if a.size > b.size
  a.chars.map { |e|
    b.sub! e, ""
  }
  a + ?_ + b
}

61 byte (jika argumennya adalah array string)

->a{a.sort_by!(&:size);a[0].chars.map{|c|a[1].sub!c,""};a*?_‌​}

Terima kasih, m-chrzan !

PHP, 154 Bytes

list($f,$s)=strlen($b=$argv[2])<strlen($a=$argv[1])?[$b,$a]:[$a,$b];foreach(str_split($f)as$x)$s=preg_replace("#(.*?)".$x."(.*)#","$1$2",$s);echo$f."_$s";

Selain itu $s=preg_replace("#(.*?)".$x."(.*)#","$1$2",$s);Anda juga bisa menggunakanif($z=strstr($s,$x))$s=strstr($s,$x,1).substr($z,1);

R, 161 byte

Ini ternyata jauh lebih lama dari yang saya harapkan, meskipun, manipulasi string biasanya membosankan di R. Saya merasa bahwa ini harus mudah golf dengan hanya menggunakan pendekatan lain.

function(x,y){s=strsplit;if(nchar(x)>nchar(y)){v=y;w=x}else{v=x;w=y};v=s(v,"")[[1]];w=s(w,"")[[1]];j=0;for(i in v){j=j+1;if(i==w[j])w[j]=""};cat(v,"_",w,sep="")}

Tidak berpakaian

function(x,y){
    s=strsplit                      # alias for stringsplit
    if(nchar(x)>nchar(y)){v=y;w=x}  # assign v/w for the short/long strings
    else{v=x;w=y}
    v=s(v,"")[[1]]                  # split short string into vector
    w=s(w,"")[[1]]                  # split long string into vector
    j=0
    for(i in v){                    # for each char in short string, check
        j=j+1                       # if is equal to corresponding char in
        if(i==w[j])w[j]=""          # long, replace long with "" if true
    }
    cat(v,"_",w,sep="")             # insert _ and print
}

Python 2, 81 72 byte

a,b=sorted(input(),key=len)
for c in a:b=b.replace(c,"",1)
print a+"_"+b

Cobalah online

Scala, 78 byte

def f(a:String,b:String):String={if(a.size>b.size)f(b,a)else
a+"_"+(b diff a)}

Penjelasan:

def f(a:String,b:String):String={ //define a method f which has two String parameters and returns a String
                                  //(because it's recursive, scala can't figure out the return type)
  if (a.size > b.size)            //make sure that a is the shorter string
    f(b, a)
  else
    a+"_"+(b diff a)              //`b diff a` removes all elements/chars of a from b
}