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”]
]

题意：找出一个字符串所有可能的回文子串组合。

思路：深度搜索所有组合，每找出一个子串，就判断一下是否是回文串，找到最后一个位置，那么就是一个符合条件的组合。

public List<List<String>> partition(String s) {
    List<List<String>> res = new ArrayList<>();
    if (s == null || s.length() == 0) {
        return res;
    }

    HashMap<String, Boolean> memory = new HashMap<>();
    dfsPalindrome(s, 0, new ArrayList<>(), res, memory);

    return res;
}

private void dfsPalindrome(String s, int start, List<String> subres, List<List<String>> res, HashMap<String, Boolean> memory) {
    if (start == s.length()) {
        res.add(new ArrayList<>(subres));
        return;
    }

    for (int end = start + 1; end <= s.length(); end++) {
        String tmp = s.substring(start, end);
        if (!isPalindrome(tmp, memory)) {
            continue;
        }
        subres.add(tmp);
        dfsPalindrome(s, end, subres, res, memory);
        subres.remove(subres.size() - 1);
    }
}

private boolean isPalindrome(String tmp, HashMap<String, Boolean> memory) {
    if (memory.containsKey(tmp)) {
        return true;
    }
    int start = 0, end = tmp.length() - 1;
    while (start < end) {
        if (tmp.charAt(start) != tmp.charAt(end)) {
            return false;
        }
        start++;
        end--;
    }
    memory.put(tmp, true);

    return true;
}