[topic] リストの並び
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Nested HiddenConsole.WriteLine(check(new[] { 825, 102, 811, 140, 812, 125, 263 })); False
Console.WriteLine(check(new[] { 824, 102, 811, 140, 810, 155, 263 })); True
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 | using System;
using System.Linq;
class Program
{
static void Main()
{
Console.WriteLine(check(new[] { 825, 102, 811, 140, 812, 125, 263 }));
Console.WriteLine(check(new[] { 824, 102, 811, 140, 810, 155, 263 }));
}
static bool check(int[] ls)
{
if (ls.Count() <= 1)
{
return true;
}
else if (ls.First() < ls.Skip(1).First())
{
return ls.Skip(1).All(n => ls.First() < n) && check(ls.Skip(1).ToArray());
}
else if (ls.First() > ls.Skip(1).First())
{
return ls.Skip(1).All(n => ls.First() > n) && check(ls.Skip(1).ToArray());
}
else
{
return false;
}
}
}
|
Squeak Smalltalk で。
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 | | check |
check := [:list |
(1 to: list size - 1) allSatisfy: [:idx |
| first second |
(first := list at: idx) = (second := list at: idx + 1) ifTrue: [false] ifFalse: [
| sym |
sym := first < second ifTrue: [#positive] ifFalse: [#negative].
(list allButFirst: idx) - first allSatisfy: [:each | each perform: sym]]]].
check value: #(1 5 4 2). "=> true "
check value: #(1 2 3). "=> true "
check value: #(825 102 811 140 812 125 263). "=> false "
check value: #(824 102 811 140 810 155 263). "=> true "
check value: #(5 4 3 2 1). "=> true "
check value: #(4 2 5 3 1). "=> false "
|
1 2 3 4 5 6 7 8 9 10 | import Data.List
check :: [Int] -> Bool
check xs = all foo $ tails xs
where
foo [] = True
foo (x:xs) = all (EQ /=) &&& ((1 >=) . length . group) $ map (compare x) xs
(&&&) :: (a -> Bool) -> (a -> Bool) -> (a -> Bool)
(p &&& q) x = if p x then q x else False
|
syat
#6428()
[
JavaScript
]
JavaScript＠Firebugコンソールで。少し意訳しました。
>>> new Array(1,5,4,2).check(); true >>> new Array(1,2,3).check(); true >>> new Array(825,102,811,140,812,125,263).check(); false >>> new Array(824,102,811,140,810,155,263).check(); true >>> new Array(5,4,3,2,1).check(); true >>> new Array(4,2,5,3,1).check(); false
1 2 3 4 5 6 7 8 9 10 11 12 13 | Array.prototype.check = function() {
var upper = Number.MAX_VALUE;
var lower = Number.MIN_VALUE;
for (var i = 0; i < this.length; i++) {
if ( (upper <= this[i]) || (this[i] <= lower) )
return false;
if (this[i] < this[i+1])
lower = this[i];
else
upper = this[i];
}
return true;
};
|
kozima
#6429()
[
Common Lisp
]
書いてみたら意外とすっきり。
1 2 3 | (defun check (list)
(loop for (x . xs) on list always
(or (every (lambda (y) (< x y)) xs) (every (lambda (y) (> x y)) xs))))
|
Haskell の方がきれいに書けますね。
1 2 3 4 | import Data.List
check (x:xs) = (all (< x) xs || all (> x) xs) && check xs
check [] = True
|
N = 50000 のときの速度を考えて実装。
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 | object Doukaku183 extends Application {
def least [A <% Ordered[A]](a: A, b: A) = if (a < b) a else b
def greatest[A <% Ordered[A]](a: A, b: A) = if (a > b) a else b
def check[A <% Ordered[A]](ls: List[A]) = {
def loop(ls: List[A], min: A, max: A): boolean = ls match {
case Nil => true
case h :: t => {
if (min <= h && h <= max) false
else loop(t, least(min, h), greatest(max, h))
}
}
ls.reverse match {
case Nil => true
case h :: t => {
loop(t, h, h)
}
}
}
{
println(check(List(1, 5, 4, 2))) // => true
println(check(List(1, 2, 3))) // => true
println(check(List(825, 102, 811, 140, 812, 125, 263))) // => false
println(check(List(824, 102, 811, 140, 810, 155, 263))) // => true
println(check(List(5, 4, 3, 2, 1))) // => true
println(check(List(4, 2, 5, 3, 1))) // => false
println(check((1 to 50000).toList)) // => true
println(check(100 :: (1 to 49999).toList)) // => false
}
}
|
[1..100]>>=pen
#6433()
[
Haskell
]
1 2 3 4 5 6 7 8 | import Maybe
check [] = True
check xs = isJust $ check' $ reverse xs where
check' (x:xs) = foldl (\b x -> b >>= f x) (Just(x,x)) xs
f x (l,h) | x < l = Just (x,h)
| x > h = Just (l,x)
| otherwise = Nothing
|
[1..100]>>=pen
#6437()
[
Haskell
]
-∞ と ∞ と foldr を使って書き直し。
1 2 3 4 5 6 7 | import Maybe
data Z = NegInf | Finite Integer | PosInf deriving (Eq,Ord)
check xs = isJust $ foldr ((=<<).f.Finite) (Just(PosInf,NegInf)) xs where
f x (l,h) | l <= x && x <= h = Nothing
| otherwise = Just (min x l, max x h)
|
syat
#6438()
[
JavaScript
]
ナイスツッコミ！
語感にやられました。（それをMIN_VALUEと言うか・・・。）
無限大に直してみました。
>>> new Array(0, 1, 2).check();
true
語感にやられました。（それをMIN_VALUEと言うか・・・。）
無限大に直してみました。
>>> new Array(0, 1, 2).check();
true
1 2 3 4 5 | < var upper = Number.MAX_VALUE;
< var lower = Number.MIN_VALUE;
---
> var upper = Number.POSITIVE_INFINITY;
> var lower = Number.NEGATIVE_INFINITY;
|
[1..100]>>=pen
#6439()
[
diff
]
整数しかこないつもりでいました。一般の型に修正。
1 2 3 | > data Z = NegInf | Finite Integer | PosInf deriving (Eq,Ord)
---
> data N a = NegInf | Finite a | PosInf deriving (Eq,Ord)
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 | let rec check = function
| [] -> true
| hd::[] -> true
| x::y::tl ->
if (x!=y) &&
((x<y && List.for_all ((<) x) tl) ||
(x>y && List.for_all ((>) x) tl))
then check (y::tl) else false;;
(*
Array.map check [|
[1; 5; 4; 2];
[1; 2; 3];
[825; 102; 811; 140; 812; 125; 263];
[824; 102; 811; 140; 810; 155; 263];
[5; 4; 3; 2; 1];
[4; 2; 5; 3; 1];
|];;
*)
|
1 2 3 4 5 6 7 8 9 10 11 12 13 | (use srfi-1)
(define (main args)
(print
(let loop ((xs (map string->number (cdr args))))
(or (null? xs)
(null? (cdr xs))
(let ((a (car xs))
(b (cadr xs)))
(and (not (= a b))
(every (cute (if (< a b) < >) a <>) (cddr xs))
(loop (cdr xs)))))))
0)
|
coze #6421() Rating0/0=0.00
数値のリストがあります。リストのサイズは N です。 ls[0]がリストの先頭要素、ls[N-1]が最後の要素です。 問題: 入力として与えられたリストが次の条件を満たすかどうかを求めてください。 for (int i = 0; i < N-1; i++) { ls[i] != ls[i+1] である ls[i] < ls[i+1] なら、ls[i] < ls[k] である(for all k = i+1 to N-1) ls[i] > ls[i+1] なら、ls[i] > ls[k] である(for all k = i+1 to N-1) } N の範囲は 3 <= N <= 50000 です。実行例を示します。 > check([1, 5, 4, 2]) True > check([1, 2, 3]) True > check([825, 102, 811, 140, 812, 125, 263]) False > check([824, 102, 811, 140, 810, 155, 263]) True > check([5, 4, 3, 2, 1]) True > check([4, 2, 5, 3, 1]) False[ reply ]