[leet code] 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"]  ]

====================

Analysis:

It's a "Sub-set" problem.  Therefore, the idea of ours is how to relate it to the solution of "sub-set problem" (DFS).

Frame of solution for "Sub-set" problem (DFS):

main method{

create result set;

call recursive function;

return result set;

}

recursive function(given String, index, result set, single result) {

if (reached the length of the single result) add single result to result set;

else{

iteration (try all the probabilities for the 1st element){

choose the first element;

pass the remaining values/sets/substring into recursive call;

remove the chosen element and try next probability;

} 

}

}

Accordingly, for our problem this time, we should 

1. Base on this basic frame and add the isPalindrome check.  

2. "reached the length of the single result" need to be checked by index. (coz we could not directly get the length of the single result, its ArrayList<String>)

3. in iteration block, we are not trying "a", "b", "c", "d" (i.e. same element length but different possible value), but trying "a", "ab", "abc", "abcd" (i.e. different element length but all include value "a").

Having the understanding above, we can implement our idea as below:

public class Solution {    public ArrayList<ArrayList<String>> partition(String s) {        ArrayList<ArrayList<String>> rs = new ArrayList<ArrayList<String>>();        helper(s, rs, 0, new ArrayList<String>());        return rs;    }        public void helper(String s, ArrayList<ArrayList<String>> rs, int index, ArrayList<String> singleRs){        if(index == s.length()) rs.add(new ArrayList<String>(singleRs));        else{            for(int i=index; i<s.length(); i++){                if(isPalindrome(s, index, i)){ // key point of this problem                    singleRs.add(s.substring(index, i+1));                    helper(s, rs, i+1, singleRs);                    singleRs.remove(singleRs.size()-1);                }            }        }    }        public boolean isPalindrome(String s, int start, int end){        while(start<=end){            if(s.charAt(start) != s.charAt(end)) return false;            start++;            end--;        }        return true;    }}