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| 1 | +/** |
| 2 | + * [131] Palindrome Partitioning |
| 3 | + * |
| 4 | + * Given a string s, partition s such that every substring of the partition is a palindrome. |
| 5 | + * |
| 6 | + * Return all possible palindrome partitioning of s. |
| 7 | + * |
| 8 | + * Example: |
| 9 | + * |
| 10 | + * |
| 11 | + * Input: "aab" |
| 12 | + * Output: |
| 13 | + * [ |
| 14 | + * ["aa","b"], |
| 15 | + * ["a","a","b"] |
| 16 | + * ] |
| 17 | + * |
| 18 | + * |
| 19 | + */ |
| 20 | +pub struct Solution {} |
| 21 | + |
| 22 | +// submission codes start here |
| 23 | + |
| 24 | +/* |
| 25 | + 记 n 个字符的回文拆分方式是 f(n) 种, 则: |
| 26 | +
|
| 27 | + f(n) = (0..n).iter().fold(0, |acc, i| { |
| 28 | + if is_palindrome(s[i..n]) { acc + f(i-1) } else { acc } |
| 29 | + }) |
| 30 | +
|
| 31 | + 按这种方式向上递推即可, 时间复杂度为 O(N^3), 空间复杂度 O(N), 显然, is_palindrome 这一步仍然有重复计算 |
| 32 | +
|
| 33 | + is_palindrome(s[i..n]) = s[i] == s[n] && is_palindrome(s[i+1..n-1]) |
| 34 | +
|
| 35 | + 存储所有 i, n 的 is_palindrome 结果, 则可以优化 is_palindrome 的时间到 O(1) |
| 36 | +
|
| 37 | + 最后的复杂度: 时间 O(N^2), 空间 O(N^2) |
| 38 | + */ |
| 39 | +impl Solution { |
| 40 | + pub fn partition(s: String) -> Vec<Vec<String>> { |
| 41 | + let s = s.chars().collect::<Vec<_>>(); |
| 42 | + if s.is_empty() { return Vec::new() } |
| 43 | + let mut palindrome_cache = vec![vec![None; s.len()]; s.len()]; |
| 44 | + let mut res: Vec<Vec<Vec<(usize, usize)>>> = Vec::with_capacity(s.len()); |
| 45 | + res.push(vec![vec![(0,1)]]); |
| 46 | + for n in 1..s.len() { |
| 47 | + let mut curr: Vec<Vec<(usize, usize)>> = Vec::new(); |
| 48 | + for i in 0..n+1 { |
| 49 | + if Solution::is_palindrome(&mut palindrome_cache, &s, i, n) { |
| 50 | + if i > 0 { |
| 51 | + for vec in res[i-1].iter() { |
| 52 | + let mut new_vec = vec.clone(); |
| 53 | + new_vec.push((i,n+1)); |
| 54 | + curr.push(new_vec); |
| 55 | + } |
| 56 | + } else { |
| 57 | + curr.push(vec![(i, n+1)]); |
| 58 | + } |
| 59 | + } |
| 60 | + } |
| 61 | + res.push(curr); |
| 62 | + } |
| 63 | + (*res[s.len()-1]).into_iter().map(|vec| { |
| 64 | + vec.iter() |
| 65 | + .map(|&range| {s[range.0..range.1].iter().collect::<String>()}) |
| 66 | + .collect::<Vec<_>>() |
| 67 | + }).collect() |
| 68 | + } |
| 69 | + |
| 70 | + fn is_palindrome(cache: &Vec<Vec<Option<bool>>>, s: &Vec<char>, i: usize, j: usize) -> bool { |
| 71 | + if j <= i { return true } |
| 72 | + if let Some(result) = cache[i][j] { |
| 73 | + result |
| 74 | + } else { |
| 75 | + s[i] == s[j] && (i + 1 > s.len() || j < 1 || Solution::is_palindrome(cache, s, i+1, j-1)) |
| 76 | + } |
| 77 | + } |
| 78 | +} |
| 79 | + |
| 80 | +// submission codes end |
| 81 | + |
| 82 | +#[cfg(test)] |
| 83 | +mod tests { |
| 84 | + use super::*; |
| 85 | + |
| 86 | + #[test] |
| 87 | + fn test_131() { |
| 88 | + assert_eq!( |
| 89 | + Solution::partition("aab".to_owned()), |
| 90 | + vec![ |
| 91 | + vec_string!["aa", "b"], |
| 92 | + vec_string!["a", "a", "b"], |
| 93 | + ]); |
| 94 | + assert_eq!( |
| 95 | + Solution::partition("aaa".to_owned()), |
| 96 | + vec![ |
| 97 | + vec_string!["aaa"], |
| 98 | + vec_string!["a", "aa"], |
| 99 | + vec_string!["aa", "a"], |
| 100 | + vec_string!["a", "a", "a"], |
| 101 | + ]); |
| 102 | + } |
| 103 | +} |
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