lintcode 容易题：Palindrome Partitioning 分割回文串

public class Solution {
    /**
     * @param s: A string
     * @return: A list of lists of string
     */
    public ArrayList<ArrayList<String>> partition(String s) {
        // write your code here
        ArrayList<ArrayList<String>> result = new ArrayList<ArrayList<String>>();
        if(s==null)
            return result;
        ArrayList<String> path = new ArrayList<String>();
        helper(s,path,0,result);
        return result;
    }
    private boolean isPalindrome(String s){
        int beg = 0;
        int end = s.length() - 1;
        while(beg<end){
            if(s.charAt(beg)!=s.charAt(end))
                return false;
            beg++;
            end--;
        }
        return true;
    }
    private void helper(String s,ArrayList<String> path,int pos,ArrayList<ArrayList<String>> result){
        if(pos==s.length()){
            result.add(new ArrayList<String>(path));
            return;
        }
        for(int i=pos+1;i<=s.length();i++){
            String prefix = s.substring(pos,i);
            if(!isPalindrome(prefix))
                continue;
            path.add(prefix);
            helper(s,path,i,result);
            path.remove(path.size()-1);
        }
    }
}