266. Palindrome Permutation.md

Solution1 (hashmap + counter)

class Solution {
    public boolean canPermutePalindrome(String s) {
        if (s == null || s.length() == 0) {
            return true;
        }
        int oddCount = 0;
        Map<Character, Integer> map = new HashMap<>();
        for (Character c : s.toCharArray()) {
            map.put(c, map.getOrDefault(c, 0) + 1);
            oddCount += map.get(c) % 2 == 0 ? -1 : 1;
        }
        return oddCount < 2;
    }
}

Solution2 (set)

class Solution {
    public boolean canPermutePalindrome(String s) {
        Set<Character> set = new HashSet<>();
        for (Character c : s.toCharArray()) {
            if (set.contains(c)) {
                set.remove(c);
            }
            else {
                set.add(c);
            }
        }
        return set.size() <= 1;
    }
}

note

• time O(n) space O(n) for both method
• The idea is to know that a palindrome string at most have one odd number of characters.