Given a string s, partition s such that every substring of the partition is a palindrome.
Return all possible palindrome partitioning of s.
Example:
Input: "aab" Output: [ ["aa","b"], ["a","a","b"] ]使用树的dfs top-down，扫边即可。维护一个path，进孩子前加我，离开孩子前删我。 关键是要找到进孩子的条件。 —— substring 是一个 回文串。
class Solution { public List<List<String>> partition(String s) { List<List<String>> result = new ArrayList<>(); List<String> path = new ArrayList<>(); dfs(s, 0, result, path); return result; } private void dfs(String s, int start, List<List<String>> result, List<String> path) { if (start == s.length()) { result.add(new ArrayList<>(path)); } for (int i = start; i < s.length(); i++) { String e = s.substring(start, i + 1); if (isPal(e)) { path.add(e); dfs(s, i + 1, result, path); path.remove(path.size() - 1); } } } private boolean isPal(String s) { int start = 0; int end = s.length() - 1; while (start <= end) { if (s.charAt(start) != s.charAt(end)) { return false; } start++; end--; } return true; } }