Longest Palindrome: ans%2==1

First Thoughts: using character counts to determine palindrome validity. retrieving counts, so either array using ascii or hashmap for O(1). can only accept 1 once, if odd count which is greater than 1, add count-1.

My Solution:

class Solution {
    public int longestPalindrome(String s) {
        boolean acceptedOdd = false;
        int maxLength = 0;
        HashMap<Character, Integer> chars = new HashMap<>();
        for (char c : s.toCharArray()) {
            chars.putIfAbsent(c, 0);
            chars.put(c, chars.get(c)+1);
        } 
        if (chars.keySet().size()==1) return s.length();
        for (char ch : chars.keySet()) {
            int div = chars.get(ch)/2;
            maxLength += div*2;
        }
        if (s.length()!=maxLength) maxLength++;
        return maxLength;
    }
}

LeetCode Version:

class Solution {
    public int longestPalindrome(String s) {
        int[] count = new int[128];
        for (char c: s.toCharArray())
            count[c]++;

        int ans = 0;
        for (int v: count) {
            ans += v / 2 * 2;
            if (ans % 2 == 0 && v % 2 == 1)
                ans++;
        }
        return ans;
    }
}

Catch Point:

1. 그 동안 홀수를 한 번도 안더했는지 더했는지 boolean flag를 통해서 확인할 수도 있지만 accumulator가 짝수인지 홀수인지를 통해 홀수를 더한 적이 있는지 확인이 가능하다.

2. 물론 알고있겠지만 ascii를 index로 사용 가능하다.