Description#
Given a string s, return all the palindromic permutations (without duplicates) of it.
You may return the answer in any order. If s has no palindromic permutation, return an empty list.
Example 1:
Input: s = "aabb"
Output: ["abba","baab"]
Example 2:
Input: s = "abc"
Output: []
Constraints:
1 <= s.length <= 16s consists of only lowercase English letters.
Solutions#
Solution 1#
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| class Solution:
def generatePalindromes(self, s: str) -> List[str]:
def dfs(t):
if len(t) == len(s):
ans.append(t)
return
for c, v in cnt.items():
if v > 1:
cnt[c] -= 2
dfs(c + t + c)
cnt[c] += 2
cnt = Counter(s)
mid = ''
for c, v in cnt.items():
if v & 1:
if mid:
return []
mid = c
cnt[c] -= 1
ans = []
dfs(mid)
return ans
|
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| class Solution {
private List<String> ans = new ArrayList<>();
private int[] cnt = new int[26];
private int n;
public List<String> generatePalindromes(String s) {
n = s.length();
for (char c : s.toCharArray()) {
++cnt[c - 'a'];
}
String mid = "";
for (int i = 0; i < 26; ++i) {
if (cnt[i] % 2 == 1) {
if (!"".equals(mid)) {
return ans;
}
mid = String.valueOf((char) (i + 'a'));
}
}
dfs(mid);
return ans;
}
private void dfs(String t) {
if (t.length() == n) {
ans.add(t);
return;
}
for (int i = 0; i < 26; ++i) {
if (cnt[i] > 1) {
String c = String.valueOf((char) (i + 'a'));
cnt[i] -= 2;
dfs(c + t + c);
cnt[i] += 2;
}
}
}
}
|
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| class Solution {
public:
int n;
vector<string> ans;
unordered_map<char, int> cnt;
vector<string> generatePalindromes(string s) {
n = s.size();
for (char c : s) ++cnt[c];
string mid = "";
for (auto& [k, v] : cnt) {
if (v & 1) {
if (mid != "") {
return ans;
}
mid += k;
}
}
dfs(mid);
return ans;
}
void dfs(string t) {
if (t.size() == n) {
ans.push_back(t);
return;
}
for (auto& [k, v] : cnt) {
if (v > 1) {
v -= 2;
dfs(k + t + k);
v += 2;
}
}
}
};
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| func generatePalindromes(s string) []string {
cnt := map[byte]int{}
for i := range s {
cnt[s[i]]++
}
mid := ""
ans := []string{}
for k, v := range cnt {
if v%2 == 1 {
if mid != "" {
return ans
}
mid = string(k)
}
}
var dfs func(t string)
dfs = func(t string) {
if len(t) == len(s) {
ans = append(ans, t)
return
}
for k, v := range cnt {
if v > 1 {
cnt[k] -= 2
c := string(k)
dfs(c + t + c)
cnt[k] += 2
}
}
}
dfs(mid)
return ans
}
|
