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"]
  ]

采用DP来做。为了不重复计算palindrome， 使用一个map 存储从[i,j]是否为palindrome， 是为1， 不是为-1，没有测试过为0.

public class Solution {
    ArrayList<ArrayList<String>> result = null;
    int[][] map = null;
    public ArrayList<ArrayList<String>> partition(String s) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        result = new ArrayList<ArrayList<String>>();
        map = new int[s.length()][s.length()]; 
        getPartition(s, 0, new ArrayList<String>());
        return result;
    }
    public void getPartition(String s, int pos, ArrayList<String> row){
        if(pos == s.length()) result.add(row);
        for(int i = pos; i < s.length(); i ++){
            if(map[pos][i] == 0){
                if(checkPartition(s, pos, i)) map[pos][i] = 1;
                else map[pos][i] = -1;
            }
            if(map[pos][i] == 1){
                row.add(s.substring(pos, i + 1));
                getPartition(s, i + 1, new ArrayList<String>(row));
                row.remove(row.size() - 1);
            }
        }
    }
    
    public boolean checkPartition(String s, int start, int end){
        for(int i = 0; i < (end - start + 1) / 2; i ++){
            if(s.charAt(start + i) != s.charAt(end - i)) return false;
        }
        return true;
    }
}