Leetcode: Palindrome Partitioning

Given a string s, partition s such that every substring of the partition is a palindrome.

Return all possible palindrome partitioning of s.

For example, given s = "aab",
Return

  [
    ["aa","b"],
    ["a","a","b"]
  ]

难度：98，与Word Break II问题一样，NP（枚举）与DP结合的问题。参考了网上的做法，这道题是求一个字符串中回文子串的切割，并且输出切割结果，其实是Word Break II和Longest Palindromic Substring结合，该做的我们都做过了。首先我们根据Longest Palindromic Substring中的方法建立一个字典，得到字符串中的任意子串是不是回文串的字典。接下来就跟Word Break II一样，根据字典的结果进行切割，然后按照循环处理递归子问题的方法，如果当前的子串满足回文条件，就递归处理字符串剩下的子串。如果到达终点就返回当前结果。算法的复杂度跟Word Break II一样，取决于结果的数量，最坏情况是指数量级的。

这里建立字典以方便查找String s中任意substring是不是回文，用到了DP，用法很精髓。

public class Solution {
    public ArrayList<ArrayList<String>> partition(String s) {
        ArrayList<ArrayList<String>> partitions = new ArrayList<ArrayList<String>>();
        if (s == null || s.length() == 0) {
            return partitions;
        }
        boolean[][] dic = getdict(s);
        ArrayList<String> partition = new ArrayList<String>();
        helper(s, dic, 0, partition, partitions);
        return partitions;
    }
    
    public void helper(String s, boolean[][] dic, int starter, ArrayList<String> partition, ArrayList<ArrayList<String>> partitions) {
        if (starter == s.length()) {
            partitions.add(new ArrayList<String>(partition));
            return;
        }
        for (int j=starter; j<s.length(); j++) {
            if (dic[starter][j]) {
                partition.add(s.substring(starter, j+1));
                helper(s, dic, j+1, partition, partitions);
                partition.remove(partition.size() - 1);
            }
        }
    }
    
    public boolean[][] getdict(String s) {
        boolean[][] dic = new boolean[s.length()][s.length()];
        for (int i=s.length()-1; i>=0; i--) {
            for (int j=i; j<s.length(); j++) {
                if ((s.charAt(i) == s.charAt(j)) && ((j-i<=2) || dic[i+1][j-1])) {
                    dic[i][j] = true;
                }
            }
        }
        return dic;
    }
}