| 1 | // anagram_solution2.rs |
| 2 | |
| 3 | fn anagram_solution2(s1: &str, s2: &str) -> bool { |
| 4 | if s1.len() != s2.len() { |
| 5 | return false; |
| 6 | } |
| 7 | |
| 8 | // s1 和 s2 中的字符分别加入 vec_a, vec_b |
| 9 | let mut vec_a = Vec::new(); |
| 10 | let mut vec_b = Vec::new(); |
| 11 | for c in s1.chars() { vec_a.push(c); } |
| 12 | for c in s2.chars() { vec_b.push(c); } |
| 13 | |
| 14 | // pos1、pos2 索引字符 |
| 15 | let mut pos1: usize = 0; |
| 16 | let mut pos2: usize; |
| 17 | |
| 18 | // 乱序字符串标示、控制循环 |
| 19 | let mut is_anagram = true; |
| 20 | |
| 21 | // 标示字符是否在 s2 中 |
| 22 | let mut found: bool; |
| 23 | |
| 24 | while pos1 < s1.len() && is_anagram { |
| 25 | pos2 = 0; |
| 26 | found = false; |
| 27 | while pos2 < vec_b.len() && !found { |
| 28 | if vec_a[pos1] == vec_b[pos2] { |
| 29 | found = true; |
| 30 | } else { |
| 31 | pos2 += 1; |
| 32 | } |
| 33 | } |
| 34 | |
| 35 | // 某字符存在于 s2 中，将其替换成 ' ' 避免再次比较 |
| 36 | if found { |
| 37 | vec_b[pos2]= ' '; |
| 38 | } else { |
| 39 | is_anagram = false; |
| 40 | } |
| 41 | |
| 42 | // 处理 s1 中下一个字符 |
| 43 | pos1 += 1; |
| 44 | } |
| 45 | |
| 46 | is_anagram |
| 47 | } |
| 48 | |
| 49 | fn main() { |
| 50 | let s1 = "rust"; |