Rust:codewars sum by Factors_rust

Source: Internet
Author: User
Tags abs

Algorithm requirements:
For an array, such as [12,15], first the mass factorization of each element,
For example, 12 =2*2*3, with a mass factor of 2, 3.
15 =3*5, with a mass factor of 3, 5.
The array has different mass factor 2,3,5.
Then, according to the order from small to large: 2 of the quality factor has 12;3 of the quality factor has: 12, 15;5 The quality factor has: 15.
You can output [(2,12), (3,12+15), (5,15)] => simplification [(2,12), (3,27), (5,15)].

Note that the mass factor of the negative integer is the same as the mass factor of the corresponding positive integer.

First, my solution

Use Std::collections::{hashmap, hashset}; fn sum_of_divided (l:vec<i64>)-> vec< (i64, i64) > {//Your code//10-6//let mut
    t; (i64, i64) > = Vec::new (); 2: There are 2 quality factors and, 3: There are 3 quality factors and println!
    ("l:{:?}", L);

    Let Mut data:hashmap<i64, vec<i64>> = Hashmap::new (); L.clone (). Iter_mut (). Map (|x| {prime_factors (*x). ABS ())-ITER (). Map (|w|
                    {if Data.contains_key (w) {Data.get_mut (W). Unwrap (). push (*X); else {Data.insert (*w, vec![
                    *X]);

        } W}) .collect::<vec<_>> (). Len () as i64
    }) .collect::<vec<i64>> (); Let result:hashmap<i64, i64> = Data.clone (). Into_iter (). Map (| ( K, V) |
        (K,v.iter (). SUM ())
    . Collect (); Sort_vec (reSult)} fn Sort_vec (data:hashmap<i64, i64>)-> vec< (i64, i64) > {//let mut output:vec< (i64, i64) &G T
    = Vec::new ();
    Let Mut ky:vec<i64> = Data.keys (). Into_iter (). Map (|&x| x). Collect ();
    Ky.sort (); Ky.into_iter (). Map (|x| (x, * (Data.get (&x). Unwrap ())). collect::<vec< (i64, i64) >> ()} fn is_prime (n:i64)-> bool {! (
        2. (n). ABS ()). Any (|x| n% x = = 0)} fn prime_factors (n:i64)-> vec<i64> {//Your code if Is_prime (n) { Return vec!
    [n];
    Let mut m = N.abs ();
    Let Mut output:vec<i64> = Vec::new ();
        For I-2..n.abs () + 1 {if m < I {break;
        } if!is_prime (i) {continue;
            While m% i = = 0i64 {if!output.contains (&i) {Output.push (i);
        } m = m/i;
    } output.sort (); Output//println!
("output:{:?}", output); }

Second, the wonderful solution

1,

Use Std::collections::{btreeset, hashset};

fn prime_factors (mut n:i64)-> hashset<i64> {Let
    mut i = 2;
    Let Mut res = Hashset::new ();
    While I <= n/i {while
        n% i = = 0 {
            Res.insert (i);
            n/= i;
        }
        i + 1;
    }
    If n > 1 {
        res.insert (n);
    }
    Res
}

fn sum_of_divided (l:vec<i64>)-> vec< (i64, i64) > {
    l.iter ()
        . Flat_map (| &x| Prime_factors (X.abs ())
        .collect::<btreeset<_>> (). Into_iter ().
        Map (|p:i64| (P, L.iter (). cloned (). filter (|x| x p = = 0). sum ()). Collect ()
}

2,

fn Is_prime (&x: &i64)-> bool {
  (2 ...).  Take_while (|p| p*p <= x). All (|p| x P!= 0)
}

fn sum_of_divided (l:vec<i64>)-> vec< (i64, i64) > {let
  mx = l.iter (). Map (|&x| x.abs ()). Max (). Unwrap_or (0);
  (2.)
    . Take_while (|&p| p <= mx)
    . Filter (is_prime).
    Filter_map (|p| {Let
      mut nums = L.iter (). cloned (). Filter (|x| x.abs ()% P = 0). peekable ();
      Nums.peek (). cloned (). Map (|_| (p, {nums}.sum ())
    })
    . Collect ()
}

3,

 fn sum_of_divided (l:vec<i64>)-> vec< (i64, i64) > {if L.len () = = 0 {return vec![]}
    Let Max = L.iter (). Max_by_key (|x| x.abs ()). Unwrap (). ABS (); (2i64.). Filter (|&x| is_prime (x)). Take_while (|&x| x <= max). Fold (Vec::new (), |mut acc, prime| {if L.iter (). Filter (|&i| i% prime = 0). Count () > 0 {acc.push (Prime, L.iter (). Filter (|& i|
        I% Prime = = 0). sum ());
    "ACC})} fn Is_prime (n:i64)-> bool {if n = 2 {return true;}
    If n < 3 {return false;}
    If n% 2 = 0 {return false;}
    Let Sqrt_limit = (n as f64). sqrt () as i64;
    Let mut i:i64 = 3;   
       While I <= sqrt_limit {if n% i = = 0 {return false;}
    i + 2; } true} 

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