(s1: &str, s2: &str)
| 1 | fn anagram_solution4(s1: &str, s2: &str) -> bool { |
| 2 | if s1.len() != s2.len() { return false; } |
| 3 | |
| 4 | // 大小为 26 的集合，用于将字符映射为 ASCII 值 |
| 5 | let mut c1 = [0; 26]; |
| 6 | let mut c2 = [0; 26]; |
| 7 | for c in s1.chars() { |
| 8 | let pos = (c as usize) - 97; // 97 为 a 的 ASCII 值 |
| 9 | c1[pos] += 1; |
| 10 | } |
| 11 | for c in s2.chars() { |
| 12 | let pos = (c as usize) - 97; |
| 13 | c2[pos] += 1; |
| 14 | } |
| 15 | |
| 16 | // 逐个比较 ascii 值 |
| 17 | let mut pos = 0; |
| 18 | let mut ok = true; |
| 19 | while pos < 26 && ok { |
| 20 | if c1[pos] == c2[pos] { |
| 21 | pos += 1; |
| 22 | } else { |
| 23 | ok = false; |
| 24 | } |
| 25 | } |
| 26 | |
| 27 | ok |
| 28 | } |
| 29 | |
| 30 | fn main() { |
| 31 | let s1 = "rust"; |