确定是否一词,或短语是等值线(isograms).

在等值线(又称为”无定形的 Word 是一个字或短语”)没有重复字母的单词或短语的逻辑术语，但允许连字符和空格出现多次。

等值线的例子:

• lumberjacks
• background
• downstream
• six-year-old

检测 isograms 单词，但它不是等值线,因为s重复了.

维基百科https://en.wikipedia.org/wiki/Isogram

pub fn check(candidate: &str) -> bool {
   unimplemented!("Is {} an isogram?", candidate);
}

# #![allow(unused_variables)]
#fn main() {
#[test]
fn empty_string() {
   assert_eq!(check(""), true, "An empty string should be an isogram.")
}

#[test]
//#[ignore]
fn only_lower_case_characters() {
   assert_eq!(check("isogram"), true, "\"isogram\" should be an isogram.")
}

#[test]
//#[ignore]
fn one_duplicated_character() {
   assert_eq!(
       check("eleven"),
       false,
       "\"eleven\" has more than one \'e\', therefore it is no isogram."
   )
}

#[test]
//#[ignore]
fn longest_reported_english_isogram() {
   assert_eq!(
       check("subdermatoglyphic"),
       true,
       "\"subdermatoglyphic\" should be an isogram."
   )
}

#[test]
//#[ignore]
fn one_duplicated_character_mixed_case() {
   assert_eq!(
       check("Alphabet"),
       false,
       "\"Alphabet\" has more than one \'a\', therefore it is no isogram."
   )
}

#[test]
//#[ignore]
fn hypothetical_isogramic_word_with_hyphen() {
   assert_eq!(
       check("thumbscrew-japingly"),
       true,
       "\"thumbscrew-japingly\" should be an isogram."
   )
}

#[test]
//#[ignore]
fn isogram_with_duplicated_hyphen() {
   assert_eq!(
       check("six-year-old"),
       true,
       "\"six-year-old\" should be an isogram."
   )
}

#[test]
//#[ignore]
fn made_up_name_that_is_an_isogram() {
   assert_eq!(
       check("Emily Jung Schwartzkopf"),
       true,
       "\"Emily Jung Schwartzkopf\" should be an isogram."
   )
}

#[test]
//#[ignore]
fn duplicated_character_in_the_middle() {
   assert_eq!(
       check("accentor"),
       false,
       "\"accentor\" has more than one \'c\', therefore it is no isogram."
   )
}

#}

# #![allow(unused_variables)]
#fn main() {
pub fn check(word: &str) -> bool {
   // Filter all non-Alphabetic character out and collect them in a new String
   let normalized_string: String = word.to_lowercase()
       .chars()
       .filter(|c| c.is_alphabetic())
       .collect();

   /* Find the char element from back and front and compare the index.
      If it is the same unique char the index will be the same.*/
   let is_unique = |x: char, word: &str| word.find(x).unwrap() == word.rfind(x).unwrap();

   // Length should be the same if it is a isogram
   normalized_string.len()
       == normalized_string
           .chars()
           .filter(|&x| is_unique(x, normalized_string.as_str()))
           .count()
}

#}