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| 1 | +pub fn manacher(s: String) -> String { |
| 2 | + let l = s.len(); |
| 3 | + if l <= 1 { |
| 4 | + return s; |
| 5 | + } |
| 6 | + |
| 7 | + // MEMO: We need to detect odd palindrome as well, |
| 8 | + // therefore, inserting dummy string so that |
| 9 | + // we can find a pair with dummy center character. |
| 10 | + let mut chars: Vec<char> = Vec::with_capacity(s.len() * 2 + 1); |
| 11 | + for c in s.chars() { |
| 12 | + chars.push('#'); |
| 13 | + chars.push(c); |
| 14 | + } |
| 15 | + chars.push('#'); |
| 16 | + |
| 17 | + // List: storing the length of palindrome at each index of string |
| 18 | + let mut length_of_palindrome = vec![1usize; chars.len()]; |
| 19 | + // Integer: Current checking palindrome's center index |
| 20 | + let mut current_center: usize = 0; |
| 21 | + // Integer: Right edge index existing the radius away from current center |
| 22 | + let mut right_from_current_center: usize = 0; |
| 23 | + |
| 24 | + for i in 0..chars.len() { |
| 25 | + // 1: Check if we are looking at right side of palindrome. |
| 26 | + if right_from_current_center > i && i > current_center { |
| 27 | + // 1-1: If so copy from the left side of palindrome. |
| 28 | + // If the value + index exceeds the right edge index, we should cut and check palindrome later #3. |
| 29 | + length_of_palindrome[i] = std::cmp::min( |
| 30 | + right_from_current_center - i, |
| 31 | + length_of_palindrome[2 * current_center - i], |
| 32 | + ); |
| 33 | + // 1-2: Move the checking palindrome to new index if it exceeds the right edge. |
| 34 | + if length_of_palindrome[i] + i >= right_from_current_center { |
| 35 | + current_center = i; |
| 36 | + right_from_current_center = length_of_palindrome[i] + i; |
| 37 | + // 1-3: If radius exceeds the end of list, it means checking is over. |
| 38 | + // You will never get the larger value because the string will get only shorter. |
| 39 | + if right_from_current_center >= chars.len() - 1 { |
| 40 | + break; |
| 41 | + } |
| 42 | + } else { |
| 43 | + // 1-4: If the checking index doesn't exceeds the right edge, |
| 44 | + // it means the length is just as same as the left side. |
| 45 | + // You don't need to check anymore. |
| 46 | + continue; |
| 47 | + } |
| 48 | + } |
| 49 | + |
| 50 | + // Integer: Current radius from checking index |
| 51 | + // If it's copied from left side and more than 1, |
| 52 | + // it means it's ensured so you don't need to check inside radius. |
| 53 | + let mut radius: usize = (length_of_palindrome[i] - 1) / 2; |
| 54 | + radius += 1; |
| 55 | + // 2: Checking palindrome. |
| 56 | + // Need to care about overflow usize. |
| 57 | + while i >= radius && i + radius <= chars.len() - 1 && chars[i - radius] == chars[i + radius] |
| 58 | + { |
| 59 | + length_of_palindrome[i] += 2; |
| 60 | + radius += 1; |
| 61 | + } |
| 62 | + } |
| 63 | + |
| 64 | + // 3: Find the maximum length and generate answer. |
| 65 | + let center_of_max = length_of_palindrome |
| 66 | + .iter() |
| 67 | + .enumerate() |
| 68 | + .max_by_key(|(_, &value)| value) |
| 69 | + .map(|(idx, _)| idx) |
| 70 | + .unwrap(); |
| 71 | + let radius_of_max = (length_of_palindrome[center_of_max] - 1) / 2; |
| 72 | + let answer = &chars[(center_of_max - radius_of_max)..(center_of_max + radius_of_max + 1)] |
| 73 | + .iter() |
| 74 | + .collect::<String>(); |
| 75 | + answer.replace("#", "") |
| 76 | +} |
| 77 | + |
| 78 | +#[cfg(test)] |
| 79 | +mod tests { |
| 80 | + use super::manacher; |
| 81 | + |
| 82 | + #[test] |
| 83 | + fn get_longest_palindrome_by_manacher() { |
| 84 | + assert_eq!(manacher("babad".to_string()), "aba".to_string()); |
| 85 | + assert_eq!(manacher("cbbd".to_string()), "bb".to_string()); |
| 86 | + assert_eq!(manacher("a".to_string()), "a".to_string()); |
| 87 | + |
| 88 | + let ac_ans = manacher("ac".to_string()); |
| 89 | + assert!(ac_ans == "a".to_string() || ac_ans == "c".to_string()); |
| 90 | + } |
| 91 | +} |
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