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.

266. Palindrome Permutation 的拓展，266只是判斷全排列是否存在回文的，此題要返回所有的回文全排列。提示：如果回文全排列存在，只需要生成前半段字符串，後面的直接根據前半段得到。用Permutations II or Next Permutation的方法去生成不同的全排列。

Python:

class Solution(object):
    def generatePalindromes(self, s):
        """
        :type s: str
        :rtype: List[str]
        """
        cnt = collections.Counter(s)
        mid = ‘‘.join(k for k, v in cnt.iteritems() if v % 2)
        chars = ‘‘.join(k * (v / 2) for k, v in cnt.iteritems())
        return self.permuteUnique(mid, chars) if len(mid) < 2 else []
    
    def permuteUnique(self, mid, nums):
        result = []
        used = [False] * len(nums)
        self.permuteUniqueRecu(mid, result, used, [], nums)
        return result
    
    def permuteUniqueRecu(self, mid, result, used, cur, nums):
        if len(cur) == len(nums):
            half_palindrome = ‘‘.join(cur)
            result.append(half_palindrome + mid + half_palindrome[::-1])
            return
        for i in xrange(len(nums)):
            if not used[i] and not (i > 0 and nums[i-1] == nums[i] and used[i-1]):
                used[i] = True
                cur.append(nums[i])
                self.permuteUniqueRecu(mid, result, used, cur, nums)
                cur.pop()
                used[i] = False

Python:

class Solution2(object):
    def generatePalindromes(self, s):
        """
        :type s: str
        :rtype: List[str]
        """
        cnt = collections.Counter(s)
        mid = tuple(k for k, v in cnt.iteritems() if v % 2)
        chars = ‘‘.join(k * (v / 2) for k, v in cnt.iteritems())
        return [‘‘.join(half_palindrome + mid + half_palindrome[::-1])                 for half_palindrome in set(itertools.permutations(chars))] if len(mid) < 2 else []

C++:

class Solution {
public:
    vector<string> generatePalindromes(string s) {
        vector<string> res;
        unordered_map<char, int> m;
        string t = "", mid = "";
        for (auto a : s) ++m[a];
        for (auto it : m) {
            if (it.second % 2 == 1) mid += it.first;
            t += string(it.second / 2, it.first);
            if (mid.size() > 1) return {};
        }
        permute(t, 0, mid, res);
        return res;
    }
    void permute(string &t, int start, string mid, vector<string> &res) {
        if (start >= t.size()) {
            res.push_back(t + mid + string(t.rbegin(), t.rend()));
        } 
        for (int i = start; i < t.size(); ++i) {
            if (i != start && t[i] == t[start]) continue;
            swap(t[i], t[start]);
            permute(t, start + 1, mid, res);
            swap(t[i], t[start]);
        }
    }
};

C++:

class Solution {
public:
    vector<string> generatePalindromes(string s) {
        vector<string> res;
        unordered_map<char, int> m;
        string t = "", mid = "";
        for (auto a : s) ++m[a];
        for (auto &it : m) {
            if (it.second % 2 == 1) mid += it.first;
            it.second /= 2;
            t += string(it.second, it.first);
            if (mid.size() > 1) return {};
        }
        permute(t, m, mid, "", res);
        return res;
    }
    void permute(string &t, unordered_map<char, int> &m, string mid, string out, vector<string> &res) {
        if (out.size() >= t.size()) {
            res.push_back(out + mid + string(out.rbegin(), out.rend()));
            return;
        } 
        for (auto &it : m) {
            if (it.second > 0) {
                --it.second;
                permute(t, m, mid, out + it.first, res);
                ++it.second;
            }
        }
    }
};

C++:

class Solution {
public:
    vector<string> generatePalindromes(string s) {
        vector<string> res;
        unordered_map<char, int> m;
        string t = "", mid = "";
        for (auto a : s) ++m[a];
        for (auto it : m) {
            if (it.second % 2 == 1) mid += it.first;
            t += string(it.second / 2, it.first);
            if (mid.size() > 1) return {};
        }
        sort(t.begin(), t.end());
        do {
            res.push_back(t + mid + string(t.rbegin(), t.rend()));
        } while (next_permutation(t.begin(), t.end()));
        return res;
    }
};

類似題目：

[LeetCode] 266. Palindrome Permutation 回文全排列　　

[LeetCode] 46. Permutations 全排列

[LeetCode] 47. Permutations II 全排列 II

[LeetCode] 31. Next Permutation 下一個排列

[LeetCode] 267. Palindrome Permutation II 回文全排列 II