LeetCode 题解(242) : Palindrome Permutation II

题目：

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 orNext Permutation.

题解：

按照题目里的提示一步步做即可。

C++版：

class Solution {public:    vector<string> generatePalindromes(string s) {        vector<string> results;        if(s.length() == 0)            return results;                unordered_map<char, int> d;        for(auto i : s) {            if(d.find(i) != d.end())                d[i]++;            else                d.insert(pair<char, int>(i, 1));        }                bool already = false;        string candidate = "";        string single = "";                for(auto i = d.begin(); i != d.end(); i++) {            int num = i->second / 2;            for(int j = 0; j < num; j++)                candidate += i->first;            if(i->second % 2 != 0) {                if(already)                    return results;                else {                    already = true;                    single += i->first;                }            }        }                if(candidate.length() == 0 && single.length() != 0) {            results.push_back(single);            return results;        }        string left = "";        int l = candidate.length();        recursion(left, candidate, single, l, results);        return results;    }        void recursion(string left, string candidate, string &single, int &l, vector<string> &results) {        if(left.length() == l) {            string result = left + single;            reverse(left.begin(), left.end());            result += left;            results.push_back(result);            return;        }        for(int i = 0; i < candidate.length(); i++) {            if(i > 0  && candidate[i] == candidate[i - 1])                continue;            recursion(left + candidate[i], candidate.substr(0, i) + candidate.substr(i + 1), single, l, results);        }    }};

Java版：

public class Solution {    public List<String> generatePalindromes(String s) {        List<String> results = new ArrayList<>();        if(s.length() == 0)            return results;                    HashMap<Character, Integer> d = new HashMap<>();        for(int i = 0; i < s.length(); i++) {            if(d.containsKey(s.charAt(i)))                d.put(s.charAt(i), d.get(s.charAt(i)) + 1);            else                d.put(s.charAt(i), 1);        }                String candidate = "";        String single = "";        boolean already = false;                for(Character c : d.keySet()) {            int num = d.get(c) / 2;            for(int i = 0; i < num; i++)                candidate += c;            if(d.get(c) % 2 != 0) {                if(already)                    return results;                else {                    already = true;                    single += c;                }            }        }                if(candidate.length() == 0 && single.length() != 0) {            results.add(single);            return results;        }                recursion("", candidate, single, candidate.length(), results);        return results;    }        private void recursion(String left, String candidate, String single, int l, List<String> results) {        if(left.length() == l) {            String right = new StringBuffer(left).reverse().toString();            results.add(left + single + right);        }        for(int i = 0; i < candidate.length(); i++) {            if(i > 0 && candidate.charAt(i) == candidate.charAt(i - 1))                continue;            recursion(left + candidate.charAt(i), candidate.substring(0, i) + candidate.substring(i + 1), single, l, results);        }    }}

Python版：

class Solution(object):    def __init__(self):        self.results = []        def generatePalindromes(self, s):        """        :type s: str        :rtype: List[str]        """        d = {}        for i in s:            if i not in d:                d[i] = 1            else:                d[i] += 1                        already, candidate, single = False, "", ""                for i in d:            num = d[i] / 2            for j in range(num):                candidate += i            if d[i] % 2 != 0:                if already:                    return []                else:                    already = True                    single += i                            if len(candidate) == 0 and len(single) != 0:            self.results.append(single)            return self.results                            for i in range(len(candidate)):            if  i > 0 and candidate[i] == candidate[i - 1]:                continue            self.recursion(candidate[i], candidate[:i] + candidate[i+1:], len(candidate), single)                    return self.results            def recursion(self, left, candidate, l, single):        if len(left) == l:            self.results.append(left + single + left[::-1])            return        for i in range(len(candidate)):            if i > 0 and candidate[i] == candidate[i - 1]:                continue            self.recursion(left + candidate[i], candidate[:i] + candidate[i+1:], l, single)