Rust 上 2 个向量的所有可能排列

Question

下午好。我有 2 个长度不等的向量

let names = vec!["Marie", "Jana", "Eva", "Anna"];
let surnames = vec!["Novakova", "Svobodova", "Novotna", "Dvorakova", "Kralova"];

如何查看名字和姓氏的所有可能组合？

我想得到的条件3向量：

let gen_full_name = vec!["Marie Novakova", "Jana Novakova", "Eva Novotna", "Anna Novotna", "Jana Novakova", ...];

Answer 1

let names = vec!["Marie", "Jana", "Eva", "Anna"];
let surnames = vec!["Novakova", "Svobodova", "Novotna", "Dvorakova", "Kralova"];
let mut full_names = Vec::with_capacity(names.len() * surnames.len());
for name in &names {
    for surname in &surnames {
        full_names.push(format!("{} {}", name, surname));
    }
}
println!("{:?}", full_names);

无需不必要的分配。

Answer 2

这是一种使用函数式编程风格的方法。

let names = vec!["Marie", "Jana", "Eva", "Anna"];
let surnames = vec!["Novakova", "Svobodova", "Novotna", "Dvorakova", "Kralova"];
let combinations = names.iter()
    // First, map all firstnames `n` to a list containing
    // a combination of this firstname with every lastname
    .map(|&n| 
        // For every firstname, the surnames are iterated
        surnames.iter()
            // For every surname iteration, the firstname and lastname are
            // combined into a single string 
            .map(move |&s| format!("{} {}", &n, &s))
            .collect::<Vec<_>>())
    // the lists are seperated for every first name so far.
    // last but not least they must be joined together.
    .fold(vec![], |mut v, mut dat| { v.append(&mut dat); v} );