267: Palindrome Permutation II

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:

1. If a palindromic permutation exists, we just need to generate the first half of the string.

2. To generate all distinct permutations of a (half of) string, use a similar approach from: Permutations II or Next Permutation.

public class Solution {
    public List<String> generatePalindromes(String s) {
        List<String> result = new ArrayList<>();
        if (s == null || s.length() == 0) {
            return result;
        }

        // Step 1: determine if the string s is palindrome permutated
        Map<Character, Integer> map = new HashMap();
        if (!isPalindromePermutation(s, map)) {
            return result;
        }

        // Step 2: form the left half of the seed string
        StringBuffer sb = new StringBuffer();
        char middle = '\0';

        Iterator it = map.entrySet().iterator();
        while (it.hasNext()) {
            Map.Entry pair = (Map.Entry) it.next();
            char key = (char) pair.getKey();
            int val = (int) pair.getValue();
            while (val > 1) {
                sb.append(key);
                val -= 2;
            }

            if (val == 1) {
                middle = key;
            }
        }

        // Step 3: gnerate the permutations of the string
        permutation(sb.toString().toCharArray(), 0, result);
        List<String> result2 = new ArrayList<>();

        // Step 4: append the right half of the string
        for (String str : result) {
            StringBuffer tmp = new StringBuffer(str);
            if (middle != '\0') {
                tmp.append(middle);
            }

            for (int i = str.length() - 1; i >= 0; i--) {
                tmp.append(str.charAt(i));
            }
            result2.add(tmp.toString());
        }

        return result2;
    }

    private boolean isPalindromePermutation(String s, Map<Character, Integer> map) {
        int tolerance = 0;
        for (int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);
            if (map.containsKey(c)) {
                int val = map.get(c);
                map.put(c, val + 1);
            } else {
                map.put(c, 1);
            }
        }

        Iterator it = map.entrySet().iterator();
        while (it.hasNext()) {
            Map.Entry pair = (Map.Entry) it.next();
            int val = (int) pair.getValue();
            if (val % 2 == 1) {
                tolerance++;
            }
        }

        if (tolerance <= 1) {
            return true;
        } else {
            return false;
        }
    }

    private void permutation(char[] s, int start, List<String> result) {
        if (start >= s.length) {
            result.add(new String(s));
            return;
        }

        for (int i = start; i < s.length; i++) {
            if (!containsDuplicate(s, start, i)) {
                swap(s, i, start);
                permutation(s, start + 1, result);
                swap(s, i, start);
            }
        }
    }

    private void swap(char[] s, int i, int j) {
        char temp = s[i];
        s[i] = s[j];
        s[j] = temp;
    }

    private boolean containsDuplicate(char[] s, int start, int end) {
        for (int i = start; i < end; i++) {
            if (s[i] == s[end]) {
                return true;
            }
        }

        return false;
    }
}