| 1 | fn anagram_solution3(s1: &str, s2: &str) -> bool { |
| 2 | if s1.len() != s2.len() { return false; } |
| 3 | |
| 4 | // s1 和 s2 中的字符分别加入 alist, blist 并排序 |
| 5 | let mut alist = Vec::new(); |
| 6 | let mut blist = Vec::new(); |
| 7 | for c in s1.chars() { alist.push(c); } |
| 8 | for c in s2.chars() { blist.push(c); } |
| 9 | alist.sort(); blist.sort(); |
| 10 | |
| 11 | // 逐个比较排序的集合，任何字符不匹配就退出循环 |
| 12 | let mut pos: usize = 0; |
| 13 | let mut ok = true; |
| 14 | while pos < alist.len() && ok { |
| 15 | if alist[pos] == blist[pos] { |
| 16 | pos += 1; |
| 17 | } else { |
| 18 | ok = false; |
| 19 | } |
| 20 | } |
| 21 | |
| 22 | ok |
| 23 | } |
| 24 | |
| 25 | fn main() { |
| 26 | let s1 = "rust"; |