Return all possible palindrome partitioning of s.
For example, given s = "aab",
Return
[
["aa","b"],
["a","a","b"]
]
[分析] 递归尝试所有的分割法。分割当前子串,第一刀依次切取长度为1,为2,一直到最后,若切取的为回文串对后面未分割的子串继续递归分割。借助辅助数组isPalin[][]在分割时剪枝。递归的思路和WordBreakII 非常相似。
public class Solution { public List<List<String>> partition(String s) { List<List<String>> result = new ArrayList<List<String>>(); if (s == null || s.length() == 0) return result; int N = s.length(); boolean[][] isPalin = new boolean[N][N]; for (int i = 0; i < N - 1; i++) { isPalin[i][i] = true; isPalin[i][i + 1] = s.charAt(i) == s.charAt(i + 1); } isPalin[N - 1][N - 1] = true; for (int len = 3; len <= N; len++) { for (int i = 0; i + len - 1 < N; i++) { int j = i + len - 1; isPalin[i][j] = s.charAt(i) == s.charAt(j) && isPalin[i + 1][j - 1]; } } recur(s, 0, isPalin, new ArrayList<String>(), result); return result; } public void recur(String s, int start, boolean[][] isPalin, List<String> item, List<List<String>> result) { if (start == s.length()) { result.add(new ArrayList<String>(item)); return; } int N = s.length(); for (int i = start; i < N; i++) { if(isPalin[start][i]) { item.add(s.substring(start, i + 1)); recur(s, i + 1, isPalin, item, result); item.remove(item.size() - 1); } } } }