carloscn
commented
11 months ago Analysis
pub fn count_asterisks(s: &str) -> i32
{
let mut ret:i32 = 0;
if s.len() < 1 {
return ret;
}
let mut i:usize = 0;
let s_vec:Vec<&str> = s.split('|').into_iter().collect();
while i < s_vec.len() {
i += 1;
if (i & 0x1) == 0 {
continue;
}
let e = s_vec[i - 1];
ret += e.chars().fold(0, |mut count, x| {
if x == '*' {
count += 1;
}
count
});
}
return ret;
}
Description
You are given a string s, where every two consecutive vertical bars '|' are grouped into a pair. In other words, the 1st and 2nd '|' make a pair, the 3rd and 4th '|' make a pair, and so forth.
Return the number of '' in s, excluding the '' between each pair of '|'.
Note that each '|' will belong to exactly one pair.
Example 1:
Input: s = "l|eet|c*o|de|" Output: 2 Explanation: The considered characters are underlined: "l|eet|c*o|de|". The characters between the first and second '|' are excluded from the answer. Also, the characters between the third and fourth '|' are excluded from the answer. There are 2 asterisks considered. Therefore, we return 2.
Example 2:
Input: s = "iamprogrammer" Output: 0 Explanation: In this example, there are no asterisks in s. Therefore, we return 0.
Example 3:
Input: s = "yo|uar|e|b|eau|tifu|l" Output: 5 Explanation: The considered characters are underlined: "yo|uar|e|b|eau|tifu|l". There are 5 asterisks considered. Therefore, we return 5.
Constraints:
1 <= s.length <= 1000 s consists of lowercase English letters, vertical bars '|', and asterisks '*'. s contains an even number of vertical bars '|'.