Palindrome Partitioning

1. Title

Palindrome Partitioning

2. Http address

https://leetcode.com/problems/palindrome-partitioning/

3. The question

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"]
  ]

4. My code (AC)

    // Accepted
    public List<List<String>> partition(String s) {
        List<List<String>> result = new ArrayList<List<String>>();
        if( s == null || s.length() <= 0)
            return result;
        int len = s.length();
        String subStr= "";
        for(int i = 1; i <= len ;i++)
        {
            List<String> subRes = new ArrayList<String>();
            subStr = s.substring(0, i);
            if( !isPalindrome(subStr) )
            {
                continue;
            }
            subRes.add(subStr);
            partitionHlper(s,i,subRes,result);
        }
       return result;
    }

    private void partitionHlper(String s, int begin, List<String> subRes,
            List<List<String>> result) {
        
        int len = s.length();
        if( begin >= len)
        {
            result.add(new ArrayList<String>(subRes));
            return;
        }
        
        String subStr ="";
        for(int i = begin + 1; i <= len; i++)
        {
            subStr = s.substring(begin, i);
            if( !isPalindrome(subStr) )
            {
                continue;
            }
            subRes.add(subStr);
            partitionHlper(s,i,subRes,result);
            subRes.remove(subRes.size() - 1);
        }
        
    }

    private boolean isPalindrome(String str) {
        // TODO Auto-generated method stub
        if( str == null || str.length() <=1)
            return true;
        int len = str.length();
        int i = 0, j = len -1;
        while( i < j)
        {
            if( str.charAt(i) != str.charAt(j))
            {
                return false;
            }
            i++;
            j--;
        }
        return true;
    }