Palindrome Permutation II

Source

• leetcode: 267

Given a string s, return all the palindromic permutations (without duplicates) of it. Return an empty list if no palindromic permutation could be form.

For example:

Given s = "aabb", return ["abba", "baab"].

Given s = "abc", return [].

Hint:

If a palindromic permutation exists, we just need to generate the first half of the string.
To generate all distinct permutations of a (half of) string, use a similar approach from: Permutations II or Next Permutation.

Java

public List<String> generatePalindromes(String s) {
    int odd = 0;
    String mid = "";
    List<String> res = new ArrayList<>();
    List<Character> list = new ArrayList<>();
    Map<Character, Integer> map = new HashMap<>();

    // step 1. build character count map and count odds
    for (int i = 0; i < s.length(); i++) {
        char c = s.charAt(i);
        map.put(c, map.containsKey(c) ? map.get(c) + 1 : 1);
        odd += map.get(c) % 2 != 0 ? 1 : -1;
    }

    // cannot form any palindromic string
    if (odd > 1) return res;

    // step 2. add half count of each character to list
    for (Map.Entry<Character, Integer> entry : map.entrySet()) {
        char key = entry.getKey();
        int val = entry.getValue();

        if (val % 2 != 0) mid += key;

        for (int i = 0; i < val / 2; i++) list.add(key);
    }

    // step 3. generate all the permutations
    getPerm(list, mid, new boolean[list.size()], new StringBuilder(), res);

    return res;
}

// generate all unique permutation from list
void getPerm(List<Character> list, String mid, boolean[] used, StringBuilder sb, List<String> res) {
    if (sb.length() == list.size()) {
        // form the palindromic string
        res.add(sb.toString() + mid + sb.reverse().toString());
        sb.reverse();
        return;
    }

    for (int i = 0; i < list.size(); i++) {
        // avoid duplication
        if (i > 0 && list.get(i) == list.get(i - 1) && !used[i - 1]) continue;

        if (!used[i]) {
            used[i] = true; sb.append(list.get(i));
            // recursion
            getPerm(list, mid, used, sb, res);
            // backtracking
            used[i] = false; sb.deleteCharAt(sb.length() - 1);
        }
    }
}