Sfida:

Prendi un vettore / elenco di numeri interi come input e genera il numero più grande adiacente a zero.

specifiche tecniche:

• Come sempre, formato di input e output opzionale
• Si può presumere che ci sarà almeno uno zero e almeno un elemento diverso da zero.

Casi test:

1 4 3 6 0 3 7 0
7

9 4 9 0 9 0 9 15 -2
9

-4 -6 -2 0 -9
-2

-11 0 0 0 0 0 -12 10
0

0 20 
20

Buona fortuna e buon golf!

Gelatina , 8 byte

ṡ2ẠÐḟS€Ṁ

Provalo online!

ṡ2            Overlapping pairs
  ẠÐḟ         Remove pairs without zeroes
     S€       Sum of each pair
       Ṁ      Maximum

MATL , 10 byte

t~5BZ+g)X>

Provalo online! Oppure verifica tutti i casi di test .

Spiegazione

Prendiamo l'input [-4 -6 -2 0 -9]come esempio.

t     % Input array. Duplicate
      %   STACK: [-4 -6 -2 0 -9],  [-4 -6 -2 0 -9]
~     % Logical negate. Replaces zeros by logical 1, and nonzeros by logical 0
      %   STACK: [-4 -6 -2 0 -9],  [0 0 0 1 0]
5B    % Push logical array [1 0 1] (5 in binary)
      %   STACK: [-4 -6 -2 0 -9], [0 0 0 1 0], [1 0 1]
Z+    % Convolution, maintaining size. Gives nonzero (1 or 2) for neighbours of
      % zeros in the original array, and zero for the rest
      %   STACK: [-4 -6 -2 0 -9], [0 0 1 0 1]
g     % Convert to logical
      %   STACK: [-4 -6 -2 0 -9], [0 0 1 0 1]
)     % Use as index into original array
      %   STACK: [-2 -9]
X>    % Maximum of array.
      %   STACK: -2
      % Implicitly display

05AB1E , 9 byte

ü‚D€P_ÏOZ

Spiegazione

ü‚         # pair up elements
  D        # duplicate
   €P      # product of each pair (0 if the pair contains a 0)
     _     # logical negate, turns 0 into 1 and everything else to 0
      Ï    # keep only the pairs containing at least 1 zero
       O   # sum the pairs
        Z  # take max

Non funziona nell'interprete online, ma funziona offline.

Haskell, 63 43 byte

f x=maximum[a+b|(a,b)<-tail>>=zip$x,a*b==0]

Grazie a @MartinEnder per 4 byte!

Pyth, 12 11 10 byte

eSsM/#0,Vt

Coppie di moduli, filtri per zero membri, ordinamenti per somma, restituisce il valore maggiore.

JavaScript (ES6), 59 57 56 byte

let f =
    
l=>l.map((n,i)=>m=l[i-1]==0|l[i+1]==0&&n>m?n:m,m=-1/0)|m

console.log(f([1, 4, 3, 6, 0, 3, 7, 0]));       // 7
console.log(f([9, 4, 9, 0, 9, 0, 9, 15, -2]));  // 9
console.log(f([-4, -6, -2, 0, -9]));            // -2
console.log(f([-11, 0, 0, 0, 0, 0, -12, 10]));  // 0
console.log(f([3, 0, 5]));                      // 5
console.log(f([28, 0, 14, 0]));                 // 28

Modifica: salvato 2 byte grazie a Huntro
Edit: salvato 1 byte grazie a ETHproductions

JavaScript (ES6), 53 byte

a=>(m=-1/0,a.reduce((l,r)=>(m=l*r||l+r<m?m:l+r,r)),m)

Perché mi piace usare reduce. Soluzione alternativa, anche 53 byte:

a=>Math.max(...a.map((e,i)=>e*a[++i]==0?e+a[i]:-1/0))

Python, 49 byte

lambda a:max(sum(x)for x in zip(a,a[1:])if 0in x)

I test sono a ideone

Comprime le coppie, somma quelle contenenti qualsiasi zero, restituisce il massimo.

Rubino, 51 byte

->a{a.each_cons(2).map{|a,b|a*b!=0?-1.0/0:a+b}.max}

uso

f=->a{a.each_cons(2).map{|a,b|a*b!=0?-1.0/0:a+b}.max}
p f[gets.split.map(&:to_i)]

PHP, 77 68 71 byte

-3 byte da anonimo, -4 e -2 da MartinEnder

preg_match_all("#(?<=\b0 )\S+|\S+(?= 0)#",$argv[1],$m);echo max($m[0]);

Corri con php -r '<code>' '<space separated values>'

Java 7, 118 105 106 byte

int d(int[]a){int i=0,m=1<<31,c;for(;++i<a.length;m=a[i]*a[i-1]==0&(c=a[i]+a[i-‌​1])>m?c:m);return m;}

13 byte salvati grazie a @cliffroot usando invece un approccio aritmetico. 1 byte aggiuntivo grazie a @mrco dopo aver scoperto un bug (il caso test aggiunto 2, 1, 0avrebbe restituito 2invece di1 ).

Codice non testato e test:

Provalo qui.

class M{
  static int c(int[] a){
    int i,
        m = a[i=0],
        c;
    for(; ++i < a.length; m = a[i] * a[i-1] == 0 & (c = a[i] + a[i - 1]) > m)
                           ? c
                           : m);
    return m;
  }

  public static void main(String[] a){
    System.out.println(c(new int[]{ 1, 4, 3, 6, 0, 3, 7, 0 }));
    System.out.println(c(new int[]{ 9, 4, 9, 0, 9, 0, 9, 15, -2 }));
    System.out.println(c(new int[]{ -4, -6, -2, 0, -9 }));
    System.out.println(c(new int[]{ -11, 0, 0, 0, 0, 0, -12, 10 }));
    System.out.println(c(new int[]{ 0, 20 }));
    System.out.println(c(new int[]{ 2, 1, 0 }));
  }
}

Produzione:

7
9
-2
0
20
1

CJam , 16 byte

q~2ew{0&},::+:e>

Provalo online!(Come una suite di test.)

Spiegazione

q~    e# Read and eval input.
2ew   e# Get all (overlapping) pairs of adjacent values.
{0&}, e# Keep only those that contain a 0.
::+   e# Sum each pair to get the other (usually non-zero) value.
:e>   e# Find the maximum.

MATLAB con Image Processing Toolbox, 32 byte

@(x)max(x(imdilate(~x,[1 0 1])))

Questa è una funzione anonima. Esempio di utilizzo per i casi di test:

>> f = @(x)max(x(imdilate(~x,[1 0 1])))
f =
  function_handle with value:
    @(x)max(x(imdilate(~x,[1,0,1])))

>> f([1 4 3 6 0 3 7 0])
ans =
     7

>> f([9 4 9 0 9 0 9 15 -2])
ans =
     9

>> f([-4 -6 -2 0 -9])
ans =
    -2

>> f([-11 0 0 0 0 0 -12 10])
ans =
     0

>> f([0 20])
ans =
    20

Dyalog APL , 14 byte

⌈/∊2(+↑⍨0∊,)/⎕

⌈/ il più grande di

∊l'appiattito (" e nlisted"

2(... a )/coppie

 + somma (zero più qualcosa è qualcosa)

 ↑⍨ preso se

 0 zero

 ∊ è membro di

 , la coppia (lett. la concatenazione del numero della mano sinistra e del numero della mano destra)

ProvaAPL online!

R, 48 47 byte

EDIT: risolto un errore grazie a @Vlo e modificato per leggere l'input da stdins, salvato un byte assegnando we saltando le parentesi.

function(v)sort(v[c(w<-which(v==0)-1,w+1)],T)[1]

v=scan();w=which(v==0);sort(v[c(w-1,w+1)],T)[1]

Spiegazione non annidata

1. Trova indici per i quali il vettore vassume i valori 0:w <- which(v == 0)
2. Crea un nuovo vettore che contiene gli indici +-1:w-1 ew+1
3. Estrai elementi che corrispondono agli indici w-1ew+1
4. Ordinare in ordine decrescente ed estrarre l'elemento pugno

Si noti che se l'ultimo o il primo elemento di vè uno zero, w+-1verrà effettivamente recuperato un indice al di fuori della lunghezza del vettore che implica che v[length(v)+1]ritorni NA. Questo in genere non è un problema, ma le max()funzioni ritornano inavvertitamente NAse ci sono delle occorrenze nel vettore a meno che non si specifichi l'opzione na.rm=T. Quindi è più breve di 2 byte ordinare ed estrarre il primo elemento piuttosto che usarlo max(), ad esempio:

max(x,na.rm=T)
sort(x,T)[1]

Mathematica, 46 43 byte

^{Salvato 3 byte grazie a @MartinEnder .}

Max[Tr/@Partition[#,2,1]~Select~MemberQ@0]&

Funzione anonima. Prende un elenco di numeri interi come input e restituisce un numero intero come output. Basato sulla soluzione Ruby.

Perl, 42 bytes

Includes +1 for -p

Give the numbers on line on STDIN

largest0.pl <<< "8 4 0 0 5 1 2 6 9 0 6"

largest0.pl:

#!/usr/bin/perl -p
($_)=sort{$b-$a}/(?<=\b0 )\S+|\S+(?= 0)/g

Julia, 56 55 Bytes

f(l)=max(map(sum,filter(t->0 in t,zip(l,l[2:end])))...)

Create tuples for neighboring values, take those tuples containing 0, sum tuple values and find maximum

Python 2, 74 Bytes

def f(x):x=[9]+x;print max(x[i]for i in range(len(x)) if 0in x[i-1:i+2:2])

Cycle through every element, if there's a 0 in the position of either the left or the right of the current element, include it in the generator, and then run it through max. We need to pad the list with some non-0 number. It'll never be included because the slice [-1:2:2] won't include anything.

T-SQL, 182 bytes

Golfed:

DECLARE @x varchar(max)='1 5 4 3 6 1 3 17 1 -8 0 -7'

DECLARE @a INT, @b INT, @ INT WHILE @x>''SELECT @a=@b,@b=LEFT(@x,z),@x=STUFF(@x,1,z,''),@=IIF(@a=0,IIF(@b<@,@,@b),IIF(@b<>0 or @>@a,@,@a))FROM(SELECT charindex(' ',@x+' ')z)z PRINT @

Ungolfed:

DECLARE @x varchar(max)='1 5 4 3 6 1 3 17 1 -8 0 -7'

DECLARE @a INT, @b INT, @ INT
WHILE @x>''
  SELECT 
   @a=@b,
   @b=LEFT(@x,z),
   @x=STUFF(@x,1,z,''),
   @=IIF(@a=0,IIF(@b<@,@,@b),IIF(@b<>0 or @>@a,@,@a))
  FROM(SELECT charindex(' ',@x+' ')z)z 
PRINT @

Fiddle

PowerShell v3+, 62 bytes

param($n)($n[(0..$n.count|?{0-in$n[$_-1],$n[$_+1]})]|sort)[-1]

A bit longer than the other answers, but a nifty approach.

Takes input $n. Then loops through the indices 0..$n.count, uses the Where-Object (|?{...}) to pull out those indices where the previous or next item in the array is 0, and feeds those back into array slice $n[...]. We then |sort those elements, and take the biggest [-1].

Examples

PS C:\Tools\Scripts\golfing> @(1,4,3,6,0,3,7,0),@(9,4,9,0,9,0,9,15,-2),@(-4,-6,-2,0,-9),@(-11,0,0,0,0,0,-12,10)|%{""+$_+" --> "+(.\largest-number-beside-a-zero.ps1 $_)}
1 4 3 6 0 3 7 0 --> 7
9 4 9 0 9 0 9 15 -2 --> 9
-4 -6 -2 0 -9 --> -2
-11 0 0 0 0 0 -12 10 --> 0

PS C:\Tools\Scripts\golfing> @(0,20),@(20,0),@(0,7,20),@(7,0,20),@(7,0,6,20),@(20,0,6)|%{""+$_+" --> "+(.\largest-number-beside-a-zero.ps1 $_)}
0 20 --> 20
20 0 --> 20
0 7 20 --> 7
7 0 20 --> 20
7 0 6 20 --> 7
20 0 6 --> 20

q, 38 bytes

{max x where 0 in'x,'(next x),'prev x}

J, 18 bytes

[:>./2(0&e.\#+/\)]

Explanation

[:>./2(0&e.\#+/\)]  Input: array A
                 ]  Identity. Get A
     2              The constant 2
      (         )   Operate on 2 (LHS) and A (RHS)
               \    Get each subarray of size 2 from A and
             +/       Reduce it using addition
           \        Get each subarray of size 2 from A and
       0&e.           Test if 0 is a member of it
            #       Filter for the sums where 0 is contained
[:>./               Reduce using max and return

Perl 6, 53 bytes

{max map ->$/ {$1 if !$0|!$2},(1,|@_,1).rotor(3=>-2)}

Expanded:

# bare block lambda with implicit signature of (*@_)
{
  max

    map

      -> $/ {           # pointy lambda with parameter ｢$/｣
                        # ( ｢$0｣ is the same as ｢$/[0]｣ )
        $1 if !$0 | !$2 # return the middle value if either of the others is false
      },

      ( 1, |@_, 1 )     # list of inputs, with added non-zero terminals
      .rotor( 3 => -2 ) # grab 3, back-up 2, repeat until less than 3 remain
}

PHP, 66 bytes

foreach($a=$argv as$k=>$v)$v||$m=max($m,$a[$k-1],$a[$k+1]);echo$m;

Pretty straightforward. Iterates over the input, and when a number is 0, it sets $m to the highest number of the 2 adjacent numbers and any previous value of $m.

Run like this (-d added for aesthetics only):

php -d error_reporting=30709 -r 'foreach($a=$argv as$k=>$v)$v||$m=max($m,$a[$k-1],$a[$k+1]);echo$m;' -- -4 -6 -2 0 -9;echo

C# 76 74 bytes

using System.Linq;i=>i.Zip(i.Skip(1),(a,b)=>a*b==0?1<<31:a+b).Max(‌​);

Explanation:

Use zip to join the array with itself but skipping the first value in the 2nd reference so that item zero joins to item one. Multiply a times b, if the result is zero, one of them must be zero and output a + b. Otherwise, output the minimum possible integer in the language. Given the assumption that we will always have a zero and a non-zero, this minimum value will never be output as the max.

Usage:

[TestMethod]
public void LargestFriend()
{
    Assert.AreEqual(7, F(new int[] { 1, 4, 3, 6, 0, 3, 7, 0 }));
    Assert.AreEqual(9, F(new int[] { 9, 4, 9, 0, 9, 0, 9, 15, -2 }));
    Assert.AreEqual(-2, F(new int[] { -4, -6, -2, 0, -9 }));
    Assert.AreEqual(0, F(new int[] { -11, 0, 0, 0, 0, 0, -12, 10 }));
    Assert.AreEqual(20, F(new int[] { 0, 20 }));
}

R, 48 54 bytes

s=scan()

w=which;max(s[c(w(s==0)+1,w(s==0)-1)],na.rm=T)

Reads vector from console input, then takes the maximum over all values adjacent to 0.

Edit: Catches NAs produced at the boundary, thanks rturnbull!

Racket 183 bytes

(λ(k)(let*((lr(λ(l i)(list-ref l i)))(l(append(list 1)k(list 1)))(m(for/list((i(range 1(sub1(length l))))
#:when(or(= 0(lr l(sub1 i)))(= 0(lr l(add1 i)))))(lr l i))))(apply max m)))

Detailed version:

(define f
 (λ(k)
    (let* ((lr (λ(l i)(list-ref l i)))
           (l (append (list 1) k (list 1)))
           (m (for/list ((i (range 1 (sub1(length l))))
                         #:when (or (= 0 (lr l (sub1 i)))
                                    (= 0 (lr l (add1 i))) ))
                (lr l i) )))
      (apply max m) )))

Testing:

(f (list 1 4 3 6 0 3 7 0))
(f (list 9 4 9 0 9 0 9 15 -2))
(f (list -4 -6 -2 0 -9))
(f (list -11 0 0 0 0 0 -12 10))
(f (list 0 20 ))

Output:

7
9
-2
0
20

C 132 bytes

Outputs using main's return code:

int main(int a,char**_){int i,m=0;_[0]=_[a]="1";for(i=1;i<a;++i){m=(*_[i-1]-48||*_[i+1]-48?m>atoi(_[i])?m:atoi(_[i]):m);}return m;}

I feel like I should be able to save a few bytes by saving one of the atoi calls, but I couldn't find an efficient way. (,t plus t= plus , plus t twice is too long). Also this technically uses undefined behaviour (setting _[a] to "1") but every compiler I know of allows it by default.

Strategy: pad the start and end of the array with 1, then loop over the internal section checking each neighbor.

PHP 69 64 bytes

Some bytes on and off from Jörg Hülsermann and Titus. = (-5)

Requires register_globals enabled. Usage: http://localhost/notnull.php?i[]=9&i[]=-5i[]=...

$x=$_GET['i'];
$y=0;
foreach($x as $j){
if($y<abs($j)){
$y=$j;
}
}
echo $y;

Golfed:

$x=$_GET['i'];$y=0;foreach($x as $j)if($y<abs($j))$y=$j;echo $y;