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.
266. Palindrome Permutation 的拓展，266只是判断全排列是否存在回文的，此题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，后面的直接根据前半段得到。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;
}
};
类似题目：
[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 下一个排列