您可以创建从元素到数字的映射，当将其应用于其中一个序列时，将对其进行排序.然后您可以将此映射应用于其他序列.之后，您可以计算对第二个序列进行气泡排序所需的交换数:(plaground)

use std::collections::HashMap;
use core::hash::Hash;
use std::collections::HashSet;

fn create_mapping<T: Eq + Hash>(seq: &[T]) -> HashMap<&T, usize> {
    let mut ret = HashMap::new();
    for (i, elem) in seq.iter().enumerate() {
        ret.insert(elem, i);
    }
    ret
}

fn dist<T: Eq + Hash>(lhs: &[T], rhs: &[T]) -> Option<usize> {
    // Distance function is not defined for differing lengths
    if lhs.len() != rhs.len() {
        return None;
    }
    let lhs_c: HashSet<_> = lhs.iter().collect();
    let rhs_c: HashSet<_> = rhs.iter().collect();
    // Distance function is not defined if an element occurs more than once in each sequence
    if lhs.len() != lhs_c.len() || rhs.len() != rhs_c.len() {
        return None;
    }
    // Distance function is not defined if the sequences don't share all elements
    if lhs_c != rhs_c {
        return None;
    } 
    // Calculate the edit distance of `lhs` and `rhs` with only adjacent transpositions.
    let mapping = create_mapping(lhs);
    let rhs_mapped: Vec<usize> = rhs.iter().map(|e|mapping[e]).collect();
    Some(bubble_sort_length(&rhs_mapped))
}

fn bubble_sort_length(seq: &[usize]) -> usize {
    let mut ret = 0;
    for i in 0..seq.len() {
        for j in (i+1)..seq.len() {
            if seq[i]>seq[j] {
                ret += 1;
            }
        }
    }
    ret
}

#[test]
fn dist_is_none_for_differing_lengths() {
    assert_eq!(None, dist(&["b", "a", "c"], &["a", "b", "c", "d"]));
}

#[test]
fn dist_is_none_with_duplicates() {
    assert_eq!(None, dist(&["b", "a", "c", "a"], &["a", "b", "c", "d"]));
    assert_eq!(None, dist(&["a", "b", "c", "d"], &["a", "b", "c", "c"]));
}

#[test]
fn dist_is_none_with_non_empty_difference() {
    assert_eq!(None, dist(&["a", "b", "c"], &["d", "e", "f"]));
}

#[test]
fn dist_bac_to_abc_is_one() {
    assert_eq!(Some(1), dist(&["b", "a", "c"], &["a", "b", "c"]));
}

#[test]
fn dist_bca_to_bac_is_one() {
    assert_eq!(Some(1), dist(&["b", "c", "a"], &["b", "a", "c"]));
}

#[test]
fn dist_cab_to_abc_is_two() {
    assert_eq!(Some(2), dist(&["c", "a", "b"], &["a", "b", "c"]));
}

#[test]
fn dist_cdab_to_abcd_is_four() {
    assert_eq!(Some(4), dist(&["c", "d", "a", "b"], &["a", "b", "c", "d"]));
}

该算法的时间复杂度为O(n²)，其中n是序列的长度.可能有更好的方法来实现bubble_sort_length功能(参见声称时间复杂度为O(n*log(n))的https://www.geeksforgeeks.org/inversion-count-in-array-using-merge-sort/)