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21-08-2019 - |
题
这是我的(代码的高尔夫球)的挑战:采取两个阵列的字节数,并确定如第二列是一个串的第一个。如果是,出的索引内容的第二列出现在第一次。如果你找不到第二列中的第一个,然后再输出-1.
例输入:{ 63, 101, 245, 215, 0 } { 245, 215 }
预期产出:2
例输入2:{ 24, 55, 74, 3, 1 } { 24, 56, 74 }
预期产出2:-1
编辑: 有人指出,bool是多余的,因此,你的所有功能都有要做的就是返回的一个int代表指数的值或-1如果没有找到。
解决方案
Ĵ
有更多的功能37个字符要比请求:它返回的列表所有的匹配的索引
I.@(([-:#@[{.>@])"_ 0(<@}."0 _~i.@#))
用法:
NB. Give this function a name i =: I.@(([-:#@[{.>@])"_ 0(<@}."0 _~i.@#)) NB. Test #1 245 215 i 63 101 245 215 0 2 NB. Test #2 - no results 24 56 74 i 24 55 74 3 1 NB. Test #3: matches in multiple locations 1 1 i 1 1 1 2 1 1 3 0 1 4 NB. Test #4: only exact substring matches 1 2 i 0 1 2 3 1 0 2 1 2 0 1 7
NB. list[0 to end], list[1 to end], list[2 to end], ...
<@}."0 _~i.@#
NB. Does the LHS completely match the RHS (truncated to match LHS)?
[-:#@[{.>@]
NB. boolean list of match/no match
([-:#@[{.>@])"_ 0(<@}."0 _~i.@#)
NB. indices of *true* elements
I.@(([-:#@[{.>@])"_ 0(<@}."0 _~i.@#))
其他提示
普通口齿不清:
(defun golf-code (master-seq sub-seq) (search sub-seq master-seq))
后记, 149 146 170 166 167 159 个字符 (在“做工作”部分):
% define data
/A [63 101 245 215 0] def
/S [245 215] def
% do the work
/d{def}def/i{ifelse}d/l S length 1 sub d/p l d[/C{dup[eq{pop -1}{dup S p
get eq{pop p 0 eq{]length}{/p p 1 sub d C}i}{p l eq{pop}if/p l d C}i}i}d
A aload pop C
% The stack now contains -1 or the position
请注意,这会找到 最后的 如果子数组多次包含,则子数组出现。
修订记录:
- 代替
false
经过[[ne
和true
经过[[eq
保存三个字符 - 删除了如果最后一个元素可能导致漏报的错误
S
出现两次A
. 。不幸的是,这个 bug 修复有 24 个字符。 - 使错误修复变得更便宜一些,节省了四个字符
- 必须再次插入空格,因为破折号是名称中的合法字符。未捕获此语法错误,因为测试用例未到达此点。
- 停止返回布尔值,因为 OP 不再需要它们。节省 8 个字符。
解释版本:
不幸的是,SO 语法荧光笔不支持 PostScript,因此可读性仍然有限。
/A [63 101 245 215 0] def
/S [245 215 ] def
/Slast S length 1 sub def % save the index of the last element of S,
% i.e. length-1
/Spos Slast def % our current position in S; this will vary
[ % put a mark on the bottom of the stack, we need this later.
/check % This function recursively removes values from the stack
% and compares them to the values in S
{
dup [
eq
{ % we found the mark on the bottom, i.e. we have no match
pop -1 % remove the mark and push the results
}
{ % we're not at the mark yet
dup % save the top value (part of the bugfix)
S Spos get
eq
{ % the top element of the stack is equal to S[Spos]
pop % remove the saved value, we don't need it
Spos 0
eq
{ % we are at the beginning of S, so the whole thing matched.
] length % Construct an array from the remaining values
% on the stack. This is the part of A before the match,
% so its length is equal to the position of the match.
% Hence we push the result and we're done.
}
{ % we're not at the beginning of S yet, so we have to keep comparing
/Spos Spos 1 sub def % decrease Spos
check % recurse
}
ifelse
}
{ % the top element of the stack is different from S[Spos]
Spos Slast eq {pop} if % leave the saved top value on the stack
% unless we're at the end of S, because in
% this case, we have to compare it to the
% last element of S (rest of the bugfix)
/Spos Slast def % go back to the end of S
check % recurse
}
ifelse
}
ifelse
}
def % end of the definition of check
A aload % put the contents of A onto the stack; this will also push A again,
% so we have to ...
pop % ...remove it again
check % And here we go!
C99
#include <string.h>
void find_stuff(void const * const array1, const size_t array1length, /* Length in bytes, not elements */
void const * const array2, const size_t array2length, /* Length in bytes, not elements */
char * bReturnBool,
int * bReturnIndex)
{
void * found = memmem(array1, array1length, array2, array2length);
*bReturnBool = found != NULL;
*bReturnIndex = *bReturnBool ? found - array1 : -1;
}
在速记和<击>位击>很多混乱:
#include <string.h>
#define f(a,b,c,d,e,f) { void * g = memmem(a, b, c, d); f = (e = !!g) ? g - a : -1; }
蟒蛇2和3, 73 68 58字
基于上 Nikhil Chelliah's 答案 皇帝。se's 答案:
>>> t=lambda l,s:''.join(map(chr,l)).find(''.join(map(chr,s)))
>>> t([63, 101, 245, 215, 0], [245, 215])
2
>>> t([24, 55, 74, 3, 1], [24, 56, 74])
-1
蟒蛇3, 41 36字
这在一定程度上要归功于 gnibbler:
>>> t=lambda l,s:bytes(l).find(bytes(s))
>>> t([63, 101, 245, 215, 0], [245, 215])
2
>>> t([24, 55, 74, 3, 1], [24, 56, 74])
-1
Haskell, 68 64字
论点的顺序作为指定的运:
import List;t l s=maybe(-1)id$findIndex id$map(isPrefixOf s)$tails l
作为 ephemient 所指出的,我们可以关的参数和减少代码中由四个字:
import List;t s=maybe(-1)id.findIndex id.map(isPrefixOf s).tails
在的Python:
def test(large, small):
for i in range(len(large)):
if large[i:i+len(small)] == small:
return i
return -1
但是,由于人们希望简洁,不优雅:
def f(l,s):
for i in range(len(l)):
if l[i:i+len(s)]==s:return i
return -1
这是75个字符,计数空白。
红宝石,使用阵列#组(41个字符体):
def bytearray_search(a,b)
(i=b.pack('C*').index(b.pack('C*')))?i:-1
end
的Perl(36个字符体,不含参数处理):
sub bytearray_search {
($a,$b) = @_;
index(pack('C*',@$a),pack('C*',@$b))
}
我觉得我欺骗,但使用Perl,这将做OP想要什么:
sub byte_substr {
use bytes;
index shift,shift
}
在Perl通常index()
适用于与字符语义字符串,但“使用字节”编译使得它使用字节segmantics代替。从手册页:
当“使用字节”是在 效果,编码被暂时忽略,并且每个串被处理 作为一系列的字节。
另外一个在Python:
def subarray(large, small):
strsmall = ' '.join([str(c).zfill(3) for c in small])
strlarge = ' '.join([str(c).zfill(3) for c in large])
pos = strlarge.find(strsmall)
return ((pos>=0), pos//4)
红宝石1.9(44B)
_=->a,b{[*a.each_cons(b.size)].index(b)||-1}
p _[[63, 101, 245, 215, 0], [245, 215]]
p _[[24, 55, 74, 3, 1], [24, 56, 74]]
goruby(29B)
_=->a,b{a.e_(b.sz).dx(b)||-1}
<强>的Python 强>:84个字符
def f(a,b):
l=[a[i:i+len(b)]for i in range(len(a))]
return b in l and l.index(b)or-1
<强>的Prolog 强>:84个字符(表示 “否”,而不是返回-1):
s(X,[]).
s([H|T],[H|U]):-s(T,U).
f(X,Y,0):-s(X,Y).
f([_|T],Y,N):-f(T,Y,M),N is M+1.
的Python oneliner在64个字符函数定义
def f(l,s): return ''.join(map(chr,l)).find(''.join(map(chr,s)))
由于我们是显式地传递字节数组我们可以变换到Python的本机字节数组str
并使用str.find
<强> Python3 36字节强>
基于Stephan202
>>> t=lambda l,s:bytes(l).find(bytes(s))
...
>>> t([63, 101, 245, 215, 0], [245, 215])
2
>>> t([24, 55, 74, 3, 1], [24, 56, 74])
-1
在的Python:
def SearchArray(input, search):
found = -1
for i in range(0, len(input) - len(search)):
for j in range(0, len(search)):
if input[i+j] == search[j]:
found = i
else:
found = -1
break
if found >= 0:
return True, found
else:
return False, -1
要测试
print SearchArray([ 63, 101, 245, 215, 0 ], [ 245, 215 ])
print SearchArray([ 24, 55, 74, 3, 1 ], [ 24, 56, 74 ])
哪些打印:
(True, 2)
(False, -1)
请注意有一个较短的解决方案,但它使用Python语言功能是无法真正便携。
在C#:
private object[] test(byte[] a1, byte[] a2)
{
string s1 = System.Text.Encoding.ASCII.GetString(a1);
string s2 = System.Text.Encoding.ASCII.GetString(a2);
int pos = s1.IndexOf(s2, StringComparison.Ordinal);
return new object[] { (pos >= 0), pos };
}
用例:
byte[] a1 = new byte[] { 24, 55, 74, 3, 1 };
byte[] a2 = new byte[] { 24, 56, 74 };
object[] result = test(a1, a2);
Console.WriteLine("{0}, {1}", result[0], result[1]); // prints "False, -1"
public class SubArrayMatch
{
private bool _IsMatch;
private int _ReturnIndex = -1;
private List<byte> _Input;
private List<byte> _SubArray;
private bool _Terminate = false;
#region "Public Properties"
public List<byte> Input {
set { _Input = value; }
}
public List<byte> SubArray {
set { _SubArray = value; }
}
public bool IsMatch {
get { return _IsMatch; }
}
public int ReturnIndex {
get { return _ReturnIndex; }
}
#endregion
#region "Constructor"
public SubArrayMatch(List<byte> parmInput, List<byte> parmSubArray)
{
this.Input = parmInput;
this.SubArray = parmSubArray;
}
#endregion
#region "Main Method"
public void MatchSubArry()
{
int _MaxIndex;
int _Index = -1;
_MaxIndex = _Input.Count - 1;
_IsMatch = false;
foreach (byte itm in _Input) {
_Index += 1;
if (_Terminate == false) {
if (SubMatch(_Index, _MaxIndex) == true) {
_ReturnIndex = _Index;
_IsMatch = true;
return;
}
}
else {
return;
}
}
}
private bool SubMatch(int BaseIndex, int MaxIndex)
{
int _MaxSubIndex;
byte _cmpByte;
int _itr = -1;
_MaxSubIndex = _SubArray.Count - 1;
_MaxSubIndex += 1;
if (_MaxSubIndex > MaxIndex) {
_Terminate = true;
return false;
}
foreach (byte itm in _SubArray) {
_itr += 1;
_cmpByte = _Input(BaseIndex + _itr);
if (!itm == _cmpByte) {
return false;
}
}
return true;
}
#endregion
}
通过Anhar侯赛因Miah “编辑:Anhar.Miah @:03/07/2009
PHP
在105 ...
function a_m($h,$n){$m=strstr(join(",",$h),join(",",$n));return$m?(count($h)-substr_count($m,",")-1):-1;}
或更明确地,
function array_match($haystack,$needle){
$match = strstr (join(",",$haystack), join(",",$needle));
return $match?(count($haystack)-substr_count($match,",")-1):-1;
}
GNU C:
int memfind(const char * haystack, size_t haystack_size, const char * needle,
size_t needle_size)
{
const char * match = memmem(haystack, hasystack_size, needle, needle_size);
return match ? match - haystack : -1;
}
ANSI C,而不库:
int memfind(const char * haystack, size_t haystack_size, const char * needle,
size_t needle_size)
{
size_t pos = 0;
for(; pos < haystack_size; ++pos)
{
size_t i = 0;
while(pos + i < haystack_size && i < needle_size &&
haystack[pos + i] == needle[i]) ++i;
if(i == needle_size) return pos;
}
return -1;
}
<强>红宝石即可。不完全是世界上最短的,但冷静,因为它是一个扩展阵列。
class Array
def contains other=[]
index = 0
begin
matched = 0
ndx = index
while other[matched] == self[ndx]
return index if (matched+1) == other.length
matched += 1
ndx += 1
end
end until (index+=1) == length
-1
end
end
puts [ 63, 101, 245, 215, 0 ].contains [245, 215]
# 2
puts [ 24, 55, 74, 3, 1 ].contains [24, 56, 74 ]
# -1
C#,列表称为 “一” 和 “b”:
Enumerable.Range(-1, a.Count).Where(n => n == -1
|| a.Skip(n).Take(b.Count).SequenceEqual(b)).Take(2).Last();
如果你不关心返回的第一个实例,你可以做:
Enumerable.Range(-1, a.Count).Last(n => n == -1
|| a.Skip(n).Take(b.Count).SequenceEqual(b));
int m(byte[]a,int i,int y,byte[]b,int j,int z){return i<y?j<z?a[i]==b[j++]?m(a,++i,y,b,j,z):m(a,0,y,b,j,z):-1:j-y;}
Java中,116个字符。有抛出。OK额外的功能一点点,所以它的推动启动条件的组装机和数组长度到调用。这样调用它:
m(byte[] substring, int substart, int sublength, byte[] bigstring, int bigstart, int biglength)
如弗雷德里克已经张贴使用STRING转换方式的代码。这里的另一种方式,它可以使用C#来完成。
jwoolard 打我吧,顺便说一句。我也使用了相同的算法,他有。这是的,我们必须使用C ++,在大学解决的问题之一。
public static bool Contains(byte[] parent, byte[] child, out int index)
{
index = -1;
for (int i = 0; i < parent.Length - child.Length; i++)
{
for (int j = 0; j < child.Length; j++)
{
if (parent[i + j] == child[j])
index = i;
else
{
index = -1;
break;
}
}
}
return (index >= 0);
}
Lisp的V1
(defun byte-array-subseqp (subarr arr)
(let ((found (loop
for start from 0 to (- (length arr) (length subarr))
when (loop
for item across subarr
for index from start below (length arr)
for same = (= item (aref arr index))
while same
finally (return same))
do (return start))))
(values (when found t) ; "real" boolean
(or found -1))))
Lisp的V2(NB,SUBSEQ创建一个复制
(defun byte-array-subseqp (subarr arr)
(let* ((alength (length arr))
(slength (length subarr))
(found (loop
for start from 0 to (- alength slength)
when (equalp subarr (subseq arr start (+ start slength)))
do (return start))))
(values (when found t)
(or found -1))))
C#:
public static object[] isSubArray(byte[] arr1, byte[] arr2) {
int o = arr1.TakeWhile((x, i) => !arr1.Skip(i).Take(arr2.Length).SequenceEqual(arr2)).Count();
return new object[] { o < arr1.Length, (o < arr1.Length) ? o : -1 };
}
在红宝石:
def subset_match(array_one, array_two)
answer = [false, -1]
0.upto(array_one.length - 1) do |line|
right_hand = []
line.upto(line + array_two.length - 1) do |inner|
right_hand << array_one[inner]
end
if right_hand == array_two then answer = [true, line] end
end
return answer
end
实施例: IRB(主):151:0> subset_match([24,55,74,3,1],[24,56,74]) => [FALSE,-1]
IRB(主):152:0> subset_match([63,101,245,215,0],[245,215]) => [真,2]
C#,与任何类型的具有相等操作:
first
.Select((index, item) =>
first
.Skip(index)
.Take(second.Count())
.SequenceEqual(second)
? index : -1)
.FirstOrDefault(i => i >= 0)
.Select(i => i => 0 ?
new { Found = true, Index = i }
:
new { Found = false, Index - 1 });
(defun golf-code (master-seq sub-seq)
(let ((x (search sub-seq master-seq)))
(values (not (null x)) (or x -1))))
Haskell中(114个字符):
import Data.List
import Data.Maybe
g a b | elem b $ subsequences a = fromJust $ elemIndex (head b) a | otherwise = -1
红宝石,我感到看到拉尔的代码后羞愧
def contains(a1, a2)
0.upto(a1.length-a2.length) { |i| return i if a1[i, a2.length] == a2 }
-1
end
下面是一个使用串比较的C#版本。它的工作原理正确,但感觉有点哈克给我。
int FindSubArray(byte[] super, byte[] sub)
{
int i = BitConverter.ToString(super).IndexOf(BitConverter.ToString(sub));
return i < 0 ? i : i / 3;
}
// 106 characters
int F(byte[]x,byte[]y){int i=BitConverter.ToString(x)
.IndexOf(BitConverter.ToString(y));return i<0?i:i/3;}
下面是执行每个单独阵列元件的一个真比较稍长的版本。
int FindSubArray(byte[] super, byte[] sub)
{
int i, j;
for (i = super.Length - sub.Length; i >= 0; i--)
{
for (j = 0; j < sub.Length && super[i + j] == sub[j]; j++);
if (j >= sub.Length) break;
}
return i;
}
// 135 characters
int F(byte[]x,byte[]y){int i,j;for(i=x.Length-y.Length;i>=0;i--){for
(j=0;j<y.Length&&x[i+j]==y[j];j++);if(j>=y.Length)break;}return i;}