267. Palindrome Permutation II

Problem:

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.

Thoughts:

``````public class Solution {
public List<String> generatePalindromes(String s) {
int[] cha = new int [256];
for (int i = 0; i < s.length(); i ++) {
cha[s.charAt(i)] += 1;
}
boolean odd = false;
int oddIndex = 0;
for (int i = 0; i < 256; i ++) {
if (cha[i] % 2 == 1) {
if (odd == true) {
return result;
}
oddIndex = i;
odd = true;
}
}

String base = "";
if (odd == true) {
base = (char)oddIndex + "";
cha[oddIndex] -= 1;
}
process(base, cha, s.length(), result);
return result;
}
private void process(String base, int[] cha, int n, List<String> result) {
if (base.length() == n) {
return;
}
for (int i = 0; i < cha.length; i ++) {
if (cha[i] > 0) {
cha[i] -= 2;
process((char)i + base + (char)i, cha, n, result);
cha[i] += 2;
}
}
}
}
``````