题目摘录：
- 给定一系列数字，使用 vector 并返回这个列表的中位数（排列数组后位于中间的值）和众数（mode，出现次数最多的值；这里哈希 map 会很有帮助）。
- 将字符串转换为 Pig Latin，也就是每一个单词的第一个辅音字母被移动到单词的结尾并增加 “ay”，所以 “first” 会变成 “irst-fay”。元音字母开头的单词则在结尾增加 “hay”（“apple” 会变成 “apple-hay”）。牢记 UTF-8 编码！
- 使用哈希 map 和 vector，创建一个文本接口来允许用户向公司的部门中增加员工的名字。例如，“Add Sally to Engineering” 或 “Add Amir to Sales”。接着让用户获取一个部门的所有员工的列表，或者公司每个部门的所有员工按照字典序排列的列表。
使用的依赖：
[dependencies]
regex = "1"
lazy_static = "1.4.0"
代码：
use std::{collections::HashMap};
use lazy_static::lazy_static;
use regex::Regex;
fn main() {
println!("Hello, world!");
let mut nums = vec![
4, 6, 23, 0, 20, 23, 16, 3, 11, 21, 16, 8, 12, 2, 18, 16, 10, 20, 20, 1, 17, 7, 19, 24, 24,
5, 1, 14, 7, 19,
];
nums.sort();
println!("{:?}", nums);
// 获取中位数
println!("middle is: {:?}", get_middle(&nums));
// 获取众数
println!("get_most_num is: {:?}", get_most_num(&nums));
// 每一个单词的第一个辅音字母被移动到单词的结尾并增加 “ay”
let words = vec!["first", "apple", "ear"];
for w in words {
println!("the word ({}) => {}", w, pig_latin(w));
}
// 用户输入员工姓名到map，并获取员工列表
company_peosons();
}
// 计算中位数
fn get_middle(nums: &Vec<i32>) -> i32 {
let length = nums.len();
if length % 2 == 1 {
return nums[length / 2];
} else {
return (nums[length / 2 - 1] + nums[length / 2]) /2
}
}
// 计算众数，可能会有多个众数，返回数组
fn get_most_num(nums: &Vec<i32>) -> Vec<i32> {
let mut map = HashMap::new();
for num in nums.iter() {
let count = map.entry(num).or_insert(0);
*count += 1;
}
// 先将map转为vec，从而对其进行排序，找出出现次数最多的那个
let mut count_vec: Vec<_> = map.iter().collect();
count_vec.sort_by(|a, b| a.1.cmp(b.1).reverse());
println!("{:?}", count_vec);
// 找出次数最多的数字
let max = count_vec[0].1;
let mut max_num:Vec<i32> = Vec::new();
// max_num.push(**(count_vec[0].0));
for one in count_vec {
if one.1 < max {
break;
}
max_num.push(**(one.0));
}
max_num
}
fn pig_latin(s: &str) -> String{
let yuanyin = vec!['a', 'e', 'i', 'u'];
let mut new_str = String::from(s);
for (i, c) in s.chars().enumerate() {
if i == 0 && yuanyin.contains(&c){
new_str = format!("{}-{}", s, "hay");
break;
} else {
if !yuanyin.contains(&c) {
if i+1 < s.len() {
let start = &s[..i];
let end = &s[(i+1)..];
new_str = format!("{}{}-{}ay", start, end, c);
break;
} else {
new_str = format!("{}-{}ay", &s[..i], c);
break;
}
}
}
}
return new_str;
}
// 公司员工信息处理
fn company_peosons() {
print_info();
let default_employee_list = get_default_employee_list();
let default_length = default_employee_list.len();
let mut input = String::new();
let mut employee_info:HashMap<String, Vec<String>> = HashMap::new();
let mut index = 0;
loop {
if index < default_length {
input = match default_employee_list.get(index) {
Some(t) => t.clone(),
None => "".to_string(),
};
index += 1;
} else {
println!("请输入你的指令（输入 help 显示帮助信息）： ");
std::io::stdin().read_line(&mut input).expect("信息输入有误，请重试！");
// println!("input = {}", input);
}
let new_input = input.trim().to_lowercase();
// println!("new_input = {}", new_input);
match new_input.as_str() {
"list" => print_all_employee(&employee_info, false), // 输出公司所有员工信息，升序
"list-reverse" => print_all_employee(&employee_info, true), // 输出公司所有员工信息，降序
"listd" => print_department_employee(&employee_info, false), // 输出公司部门员工信息，降序
"listd-reverse" => print_department_employee(&employee_info, true), // 输出公司部门员工信息，降序
"help" => print_info(),
"quit" => return,
other => {
let y_or_n = deal_add_employee(&mut employee_info, other);
if y_or_n {
println!("添加成功！");
} else{
println!("添加失败！");
}
},
}
// 将上一次输入的数据清空
input.clear();
}
}
fn print_info() {
println!("\n");
println!("公司员工信息记录");
println!("你可以输入：Add name to job 这种格式的文本来添加员工到指定职位，例如：");
println!("Add Sally to Engineering 将Sally添加到Engineering职位下");
println!("Add Amir to Sales 将Amir添加到Sales职位下");
println!("输入 list 指令可以获取公司当前所有员工信息-升序");
println!("输入 list-reverse 指令可以获取公司当前所有员工信息-降序");
println!("输入 listd 指令可以获取公司一个部门所有员工信息-升序");
println!("输入 listd-reverse 指令可以获取公司一个部门所有员工信息-降序");
println!("输入 quit 指令退出当前程序");
println!("输入 help 指令可以获取帮助信息");
println!("\n");
}
// 设置一批默认的公司员工数据
fn get_default_employee_list() -> Vec<String> {
vec![
String::from("Add Sally anny to Engineering"),
String::from("Add Wade to Engineering"),
String::from("Add Seth to Engineering"),
String::from("Add Dave to Engineering"),
String::from("Add Gilbert to Engineering"),
String::from("Add Brian to Engineering"),
String::from("Add Liam to Engineering"),
String::from("Add Claude to Engineering"),
String::from("Add Harvey to Engineering"),
String::from("Add Amir to Sales"),
String::from("Add Connor to Sales"),
String::from("Add Kingston to Sales"),
String::from("Add Harold to Sales"),
String::from("Add Eli to Sales"),
String::from("Add Carlos to Sales"),
String::from("Add Paul to Sales"),
String::from("Add Ricardo to Sales"),
String::from("Add Jessie to Manager"),
String::from("Add Wiley to Manager"),
String::from("Add Clark to Manager"),
String::from("Add Marshall to Manager"),
String::from("Add Marion to Manager"),
String::from("Add Benjamin to Manager"),
String::from("Add Terry to Manager"),
String::from("Add Joe to Manager"),
String::from("Add Christopher to Manager"),
]
}
// 处理添加的员工信息
fn deal_add_employee(employee_info: &mut HashMap<String, Vec<String>>, new_input: &str) -> bool {
lazy_static! {
static ref RE: Regex = Regex::new(r"([Aa][Dd]{2}) +(.+?) +([Tt][Oo]) +(\w+)?").unwrap();
}
for cap in RE.captures_iter(new_input) {
// println!("{:?}", &cap);
let s1 = match &cap.get(1).as_mut() {
Some(t) => t.as_str().to_lowercase(),
None => "".to_string(),
};
let s2 = match &cap.get(2).as_mut() {
Some(t) => t.as_str().to_lowercase(),
None => "".to_string(),
};
let s3 = match &cap.get(3).as_mut() {
Some(t) => t.as_str().to_lowercase(),
None => "".to_string(),
};
let s4 = match &cap.get(4).as_mut() {
Some(t) => t.as_str().to_lowercase(),
None => "".to_string(),
};
if s1 == "add" && s3 == "to" {
let vector = employee_info.entry(s4.clone())
.or_insert(Vec::new());
vector.push(s2.clone());
return true;
}
}
return false;
}
// 将公司中所有的员工合并到一起并排序
fn print_all_employee(employee_info: &HashMap<String, Vec<String>>, reverse: bool) {
let mut all_emp = Vec::new();
for emp_list in employee_info.values() {
all_emp.extend(emp_list.iter());
}
sort_vect_list(&mut all_emp, reverse);
println!("{:?}", all_emp);
}
fn print_department_employee(employee_info: &HashMap<String, Vec<String>>, reverse: bool) {
// 先获取部门值并排序
let mut department_keys = Vec::new();
for key in employee_info.keys() {
department_keys.push(key);
}
sort_vect_list(&mut department_keys, reverse);
for department in department_keys {
println!("department name: {}", department);
let mut peosons = Vec::new();
match employee_info.get(department) {
Some(value) => {
for p in value {
peosons.push(p);
}
},
None => {},
};
sort_vect_list(&mut peosons, reverse);
println!("{:?}", peosons);
println!("");
}
}
// 对vect数组进行排序
fn sort_vect_list(sort_list: &mut Vec<&String>, reverse: bool) {
if reverse {
sort_list.sort_by(|a,b| a.cmp(b).reverse());
} else {
sort_list.sort_by(|a,b| a.cmp(b));
}
}
标签：练习题,mut,Latin,String,Pig,employee,Add,let,println From： https://www.cnblogs.com/ctsch/p/17089396.html