# rust 算法题求优化之 数组交集并排字典序

## 英文原文描述：

Description:
Given two arrays of strings a1 and a2 return a sorted array r in lexicographical order of the strings of a1 which are substrings of strings of a2.

#Example 1: a1 = [“arp”, “live”, “strong”]

a2 = [“lively”, “alive”, “harp”, “sharp”, “armstrong”]

returns [“arp”, “live”, “strong”]

#Example 2: a1 = [“tarp”, “mice”, “bull”]

a2 = [“lively”, “alive”, “harp”, “sharp”, “armstrong”]

returns []

Notes:
Arrays are written in “general” notation. See “Your Test Cases” for examples in your language.

In Shell bash a1 and a2 are strings. The return is a string where words are separated by commas.

Beware: r must be without duplicates.
Don’t mutate the inputs.

## 中文简述：

``````fn in_array(arr_a: &[&str], arr_b: &[&str]) -> Vec<String> {
todo!();
}``````

``````fn in_array(arr_a: &[&str], arr_b: &[&str]) -> Vec<String> {
let mut vec:Vec<String> = Vec::new();
for &i in arr_a {
for &j in arr_b {
if j.contains(i) && !vec.contains(&String::from(i)) {
vec.push(String::from(i));
}
}
}
vec.sort();
vec
}``````

``````fn in_array(arr_a: &[&str], arr_b: &[&str]) -> Vec<String> {
let mut vec: Vec<String> = arr_a.iter().filter(|&&str1| arr_b.iter().any(|&str2| str2.contains(&str1))).map(|&s| String::from(s)).collect();
vec.sort();
vec.dedup();
vec
}``````
8个月前 评论

``````fn in_array(arr_a: &[&str], arr_b: &[&str]) -> Vec<String> {
let mut vec: Vec<String> = arr_a.iter().filter(|&&str1| arr_b.iter().any(|&str2| str2.contains(&str1))).map(|&s| String::from(s)).collect();
vec.sort();
vec.dedup();
vec
}``````
8个月前 评论

8个月前 评论