reconstruct-original-digits-from-english.rs

#![allow(dead_code, unused, unused_variables)]

fnmain(){
assert_eq!(
"012".to_string(),
Solution::original_digits("owoztneoer".to_string())
);
assert_eq!(
"0123456789".to_string(),
Solution::original_digits("zeroonetwothreefourfivesixseveneightnine".to_string())
);
}

structSolution;

implSolution{
// zero
// one
// two
// three
// four
// five
// six
// seven
// eight
// nine
// efghinorstuvwxz
pubfnoriginal_digits(s:String) -> String{
letmut l = [0;26];
letmut count = [0;10];// 用于记录每个数字的出现次数
for&i in s.as_bytes().iter(){
 l[(i - 97)asusize] += 1;
}

letmut s = String::new();

for&i in"zwuxghfsin".as_bytes().iter(){
if i == b'z'{
 count[0] += l[(i - 97)asusize]
}elseif i == b'w'{
 count[2] += l[(i - 97)asusize]
}elseif i == b'u'{
 count[4] += l[(i - 97)asusize]
}elseif i == b'x'{
 count[6] += l[(i - 97)asusize]
}elseif i == b'g'{
 count[8] += l[(i - 97)asusize]
}elseif i == b'h'{
 count[3] += l[(i - 97)asusize] - count[8]// 三的数量是h的数量剪掉8的数量
}elseif i == b'f'{
 count[5] += l[(i - 97)asusize] - count[4]
}elseif i == b's'{
 count[7] += l[(i - 97)asusize] - count[6]
}elseif i == b'i'{
 count[9] += l[(i - 97)asusize] - count[6] - count[5] - count[8]
}elseif i == b'n'{
 count[1] += l[(i - 97)asusize] - count[7] - count[9]*2// 因为nine是两个n
}
}

for(i,&v)in count.iter().enumerate(){
if v != 0{
 s.push_str(i.to_string().repeat(v).as_str());
}
}

 s
}
}