Rust : codewars的Sum of Pairs

还是见codewars的Sum of Pairs：

一、相关的要求

Given a list of integers and a single sum value, return the first two values (parse from the left please) in order of appearance that add up to form the sum.sum_pairs([11, 3, 7, 5],         10)
#              ^--^      3 + 7 = 10
== [3, 7]sum_pairs([4, 3, 2, 3, 4],         6)
#          ^-----^         4 + 2 = 6, indices: 0, 2 *
#             ^-----^      3 + 3 = 6, indices: 1, 3
#                ^-----^   2 + 4 = 6, indices: 2, 4
#  * entire pair is earlier, and therefore is the correct answer
== [4, 2]sum_pairs([0, 0, -2, 3], 2)
#  there are no pairs of values that can be added to produce 2.
== None/nil/undefined (Based on the language)sum_pairs([10, 5, 2, 3, 7, 5],         10)
#              ^-----------^   5 + 5 = 10, indices: 1, 5
#                    ^--^      3 + 7 = 10, indices: 3, 4 *
#  * entire pair is earlier, and therefore is the correct answer
== [3, 7]
Negative numbers and duplicate numbers can and will appear.NOTE: There will also be lists tested of lengths upwards of 10,000,000 elements. Be sure your code doesn't time out.

二、我的解法

use std::collections::HashMap;
fn sum_pairs(ints: &[i8], s: i8) -> Option<(i8, i8)> {// your codelet mut pair: HashMap<i64, Option<(i8, i8)>> = HashMap::new();let mut c = 0_i64;(&ints).into_iter().filter(|&x| {c += 1_i64;let mut indice = 0_64;(&ints[c as usize..]).into_iter().filter(|&y| {indice += 1_i64;match (*y) as i64  + (*x) as i64 == s as i64 {true => {pair.insert(indice + c, Some((*x, *y)));return true;}_ => return false,}}).collect::<Vec<_>>().len() > 0usize}).collect::<Vec<_>>();match pair.len() > 0 {true => {let mut indices: Vec<i64> = pair.keys().into_iter().map(|&x| x).collect();indices.sort();let min_indice = &indices.first().unwrap();//println!("pair:{:?}", pair);//println!("indices:{:?} min_indice :{:?}", indices, min_indice);return *(pair.get(&min_indice).unwrap());}_ => return None,}
}

通过了测试，问题是提交超时，有待优化。

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