Sort odd numbers first

Question

Rearrange a given list such that all the odd numbers appear before all the even numbers. Besides for this requirement, the output list may be in any order.

The input will only contain integers, but they may be negative and there may be duplicates, and they may appear in any order.

Shortest solution wins.

Test cases

[1,2] → [1,2]

[2,1] → [1,2]

[1,0,0] → [1,0,0]

[0,0,-1] → [-1,0,0]

[3,4,3] → [3,3,4]

[-4,3,3] → [3,3,-4]

[2,2,2,3] → [3,2,2,2]

[3,2,2,2,1,2] → [1,3,2,2,2,2] or [3,1,2,2,2,2]

[-2,-2,-2,-1,-2,-3] → [-1,-3,-2,-2,-2,-2,] or [-3,-1,-2,-2,-2,-2,]

[] → []

Answer 1

sed 4.2.2, 32 bytes

:
s/(<(!!)*>)(<(!!)*!>)/\3\1/g
t

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Input is formatted as a exclamation marks surrounded by <>. Output is the same.

Ex: <><!><!!><!!!><!!!!> => <!><!!!><><!!><!!!!>

Answer 2

x86-16 machine code, 25 bytes

00000000: e316 8bd7 03d1 4afc aca8 0174 03aa eb06  ......J....t....
00000010: 87d7 fdaa 87d7 e2ef c3                   .........

Listing

E3 16       JCXZ EXIT           ; handle empty input list
8B D7       MOV  DX, DI         ; DI = output list head 
03 D1       ADD  DX, CX         ; DX = output list tail (1 past)
4A          DEC  DX             ; DX = output list tail
        INLOOP: 
FC          CLD                 ; string direction forward 
AC          LODSB               ; get next char 
A8 01       TEST AL, 1          ; is even? NZ if odd, ZF if even 
74 03       JZ   EVEN 
        ODD: 
AA          STOSB               ; write to list head 
EB 06       JMP  DONE 
        EVEN: 
87 D7       XCHG DX, DI         ; DI = list tail 
FD          STD                 ; string direction reverse 
AA          STOSB               ; write to list tail 
87 D7       XCHG DX, DI         ; DI = list head 
        DONE:
E2 EF       LOOP INLOOP         ; continue looping input list
        EXIT:
C3          RET                 ; return to caller

Callable function, input list at DS:SI, length in CX. Output list buffer at ES:DI.

If input number is odd, is added to the beginning of output list. If even, is added to the end of the output list.

Test cases:

[1, 2] → [1, 2]

[0, 0, -1] → [-1, 0, 0]

[-1, -2, 3, -4, -5] → [-1, 3, -5, -4, -4]

[3, 2, 2, 2, 1, 2] → [3, 1, 2, 2, 2, 2]

Answer 3

Kotlin, 33 bytes

{l:List<Int>->l.sortedBy{1-it%2}}

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Answer 4

Swift, 22 bytes

Two different answers with the same byte count!

In-place

{$0.partition{$0&1<1}}

Argument type is any MutableCollection whose element type conforms to BinaryInteger. Technically this returns the index where it changes from odd to even numbers, but this can be made to return Void with a warning: SwiftFiddle link

C brain keeps telling me to do ~$0&1 but Swift doesn't do int-bool conversion

New array

{$0.sorted{$0&1>$1&1}}

Argument type is any Sequence whose element type conforms to BinaryInteger. Returns an array with the same element type. SwiftFiddle link

Answer 5

Arturo, 22 bytes

$=>[arrange&=>[^0-1&]]

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$=>[             ; a function where input is assigned to &
    arrange&=>[  ; sort input by
        ^0-1&    ; -1 to the (current element) power
    ]            ; end arrange
]                ; end function