Rust:字符串数组的拼接、与 String 、 &str的选择

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一、问题的由来
1、由来
我曾用Julia写过以下字符串数组的方案，可以参考：
https://blog.csdn.net/wowotuo/article/details/52740161

2、具体的问题
一个字符串数组的数组，求其字符串数组元素中的元素之间一一拼接成新字符串数组。
a = [“a1”, “a2”, “a3”];
b = [“b1”, “b2”];
c = [“c1”, “c2”];
list = [a, b, c, …];
要求根据list得到一个结果集合，类似如下格式：
[“a1b1c1”, “a1b1c2”, “a1b2c1”, “a1b2c2”, “a2b1c1”, “a2b1c2”, ……]
即结果集合的每个元素是list内每个数组取且只取一个元素拼接成的一个字符串。
需要说明的是，要提出一个抽象的解决方案，并不限于a,b,c三个数组的拼接

二、如果用Rust来写，应如何写呢？

1、方案一: 用Vec < String > 方式

use std::ops::Deref;
use std::{thread, time};
fn get_new_strings(str1: Vec<String>, str2: Vec<String>) -> Vec<String> {
    let mut strs: Vec<String> = Vec::new();
    for s in &str1 {
        for v in &str2 {
            let temp: String = (s.deref()).to_string() + v;
            strs.push(temp);
        }
    }
    strs
}

fn main() {
    let sleep_seconds = std::time::Duration::from_secs(1000);
    let mut data = Vec::new();
    let data_0: Vec<String> = ["1", "2", "3"].iter().map(|x| x.to_string()).collect();
    let data_1: Vec<String> = ["a", "b", "c", "d"].iter().map(|x| x.to_string()).collect();
    let data_2: Vec<String> = ["e", "f", "g"].iter().map(|x| x.to_string()).collect();
    let data_3: Vec<String> = ["h", "i", "j", "k"].iter().map(|x| x.to_string()).collect();
    let data_4: Vec<String> = ["x", "y", "z"].iter().map(|x| x.to_string()).collect();
    data.push(data_1);
    data.push(data_2);
    data.push(data_3);
    data.push(data_4);
    let mut temp = data_0;
    for dt in data {
        temp = get_new_strings(temp, dt);
    }
    println!("temp:{:?} count:{:?}", temp, temp.len());
    thread::sleep(sleep_seconds);
}

以上方案的不足在于输入项比较麻烦。

2、方案二 ：Vec<&str>

fn get_new_strs<'a>(str1: &Vec<&'a str>, str2: &Vec<&'a str>) -> Vec<String> {
    let mut strs: Vec<String> = Vec::new();
    for s in str1 {
        for v in str2 {
            //let temp: String = s + &v;
            let temp: String = s.to_string() + v;
            strs.push(temp);
        }
    }
    strs
}

即：

use std::ops::Deref;
use std::{thread, time};
//方案一
fn get_new_strings(str1: Vec<String>, str2: Vec<String>) -> Vec<String> {
    let mut strs: Vec<String> = Vec::new();
    for s in &str1 {
        for v in &str2 {
            //let temp: String = s + &v;
            let temp: String = (s.deref()).to_string() + v;
            strs.push(temp);
        }
    }
    strs
}
//方案二
fn get_new_strs<'a>(str1: &Vec<&'a str>, str2: &Vec<&'a str>) -> Vec<String> {
    let mut strs: Vec<String> = Vec::new();
    for s in str1 {
        for v in str2 {
            let temp: String = s.to_string() + v;
            strs.push(temp);
        }
    }
    strs
}

fn main() {
    let sleep_seconds = std::time::Duration::from_secs(1000);
    let mut data = Vec::new();
    let data_0 = vec!["1", "2", "3"];
    let data_1 = vec!["a", "b", "c", "d"];
    let data_2 = vec!["e", "f", "g"];
    let data_3 = vec!["h", "i", "j", "k"];
    let data_4 = vec!["x", "y", "z"];
    data.push(data_1);
    data.push(data_2);
    data.push(data_3);
    data.push(data_4);
    let mut outdata: Vec<String> = data_0.iter().map(|x| x.to_string()).collect();
    for _dt in data {
        //Vec<String>=>Vec<&str>
        let temp = outdata.clone();
        let _temp: Vec<&str> = temp.iter().map(|x| x.deref()).collect();
        outdata = get_new_strs(&_temp, &_dt);
    }
    println!("outdata:{:?}, count:{:?}", outdata, outdata.len());
    thread::sleep(sleep_seconds);
}

结果均一样：

三、关于String 与&str

//Vec < String > => Vec<&str>

    let mut outdata: Vec<String> = data_0.iter().map(|x| x.to_string()).collect();
    //Vec<String>=>Vec<&str>
    let temp = outdata.clone();
    let _temp: Vec<&str> = temp.iter().map(|x| x.deref()).collect();

这个可能看起来，比如对于Julia而言，是很轻松的。但是对于Rust而言，却不是轻松的。

四、其它

比如，如果能写成这样，可能是最方便的。想想看，为什么不能？

fn get_new_strs2<'a>(str1: &Vec<&'a str>, str2: &Vec<&'a str>) -> Vec<&'a str> {
    let mut strs: Vec<&'static str> = Vec::new();
    for s in str1 {
        for v in str2 {
            //let temp: String = s + &v;
            let temp: &'static str = &(s.to_string() + v);
            strs.push(temp);
        }
    }
    strs
}