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 [].
private static List<String> test(String s) {
List<String> ans = new ArrayList<>();
if (s == null || s.length() == 0) {
return ans;
}
int[] chs = new int[256];
for (int i = 0; i < s.length(); i++) {
chs[s.charAt(i)]++;
}
int count = 0;
for (int i : chs) {
if (i % 2 == 1) {
count++;
}
if (count > 1) {
return ans;
}
}
String center = "";
StringBuilder sb = new StringBuilder();
for(int i = 0; i < 256; i++) {
if (chs[i] % 2 == 1) {
center = (String.valueOf((char) i));
chs[i]--;
break;
}
}
dfs(ans, chs, center, s.length());
return ans;
}
private static void dfs(List<String> ans, int[] chs, String s, int len) {
if (s.length() == len) {
ans.add(s);
return;
}
for (int i = 0; i < 256; i++) {
if (chs[i] > 0) {
chs[i] -= 2;
dfs(ans, chs, ((char) i) + s + ((char) i), len);
chs[i] += 2;
}
}
}