Finding all palindromes in String

Finding all palindromes in String.

Naive Thinking: Use the method described in  longest-palindromic-substring . There are covered ranges for each character. That's the result we want.

Time complexity O(n^2) (because to generate output, Hoops, that seems not great benefit using longest-palindrome's algorithm) 
Space O(n^2) (Also because of the output).

 import java.util.List;  
 import java.util.ArrayList;  
 import java.util.Set;  
 import java.util.HashSet;  
 public class Solution{  
   public static void main(String args[]){  
     Solution s = new Solution();  
     List<String> list = s.findAllPalindrome("abbbbabaaabab");  
     System.out.println(list);  
   }  
   public List<String> findAllPalindrome(String s){  
     List<String> list = new ArrayList<String>();  
     Set<String> set = new HashSet<String>();  
     String str = preProcess(s);  
     int f[] = new int[str.length()]; // coverage  
     int p = 0; // pivot  
     for(int i = 1;i < str.length();i++){  
       // check if position i is in pivot's coverage  
       if(p + f[p] > i){  
         if(2*p-i - f[2*p-i] > p-f[p])  
           f[i] = f[2*p-i];  
         else  
           f[i] = f[p] - (i-p);  
       }else{  
         f[i] = 0;  
       }  
       // extend if necessary  
       int j = 1;  
       while(i-f[i]-j >= 0 && i+f[i]+j < str.length() && str.charAt(i+f[i]+j)==str.charAt(i-f[i]-j)) j++;  
       f[i] += j-1;  
       // check if need to replace pivot  
       if(i+f[i] > p+f[p]) p = i;  
     }  
     // generate result  
     for(int i = 0;i < f.length;i++)  
       for(int j = 2;j <= f[i];j++)  
         set.add(s.substring((i-j)/2,(i+j+1)/2));  
     list.addAll(set);  
     return list;  
   }  
   private String preProcess(String s){  
     StringBuilder str = new StringBuilder();  
     for(int i = 0;i < s.length();i++){  
       str.append("#");  
       str.append(s.charAt(i));  
     }  
     str.append("#");  
     return str.toString();  
   }  
 }