Given a string which consists of lowercase or uppercase letters, find the length of the longest palindromes that can be built with those letters.

This is case sensitive, for example "Aa" is not considered a palindrome here.

Note:
Assume the length of given string will not exceed 1,010.

Example:

Input:

"abccccdd"

Output:

7

Explanation:

One longest palindrome that can be built is "dccaccd", whose length is 7.

给一个含有大写和小写字母的字符串，找出能用这些字母组成的最长的回文。大小写敏感，字符长度不超过1010。

解法：由于字母可以随意组合，求出字符个数为偶数的字母，回文有两种形式，一种是左右完全对称，还有一种是以中间字符为中心，左右对称，统计出所有偶数个字符的出现总和，如果有奇数个字符的话，最后结果再加1。

Java:

public int longestPalindrome(String s) {

        if(s==null || s.length()==0) return 0;

        HashSet<Character> hs = new HashSet<Character>();

        int count = 0;

        for(int i=0; i<s.length(); i++){

            if(hs.contains(s.charAt(i))){

                hs.remove(s.charAt(i));

                count++;

            }else{

                hs.add(s.charAt(i));

            }

        }

        if(!hs.isEmpty()) return count*2+1;

        return count*2;

}

Python:

class Solution(object):

    def longestPalindrome(self, s):

        """

        :type s: str

        :rtype: int

        """

        ans = odd = 0

        cnt = collections.Counter(s)

        for c in cnt:

            ans += cnt[c]

            if cnt[c] % 2 == 1:

                ans -= 1

                odd += 1

        return ans + (odd > 0)

Python:

class Solution(object):

    def longestPalindrome(self, s):

        map = {}

        for i in s:

            if i in map:

                map[i] += 1

            else:

                map[i] = 1

        odd, even = 0, 0

        for key in map:

            if map[key] % 2:

                odd += 1

            else:

                even += 1

        return even + 1 if odd else even

Python:

import collections

class Solution(object):

    def longestPalindrome(self, s):

        """

        :type s: str

        :rtype: int

        """

        odds = 0

        for k, v in collections.Counter(s).iteritems():

            odds += v & 1

        return len(s) - odds + int(odds > 0)

    def longestPalindrome2(self, s):

        """

        :type s: str

        :rtype: int

        """

        odd = sum(map(lambda x: x & 1, collections.Counter(s).values()))

        return len(s) - odd + int(odd > 0)

C++:

class Solution {

public:

    int longestPalindrome(string s) {

        int odds = 0;

        for (auto c = 'A'; c <= 'z'; ++c) {

            odds += count(s.cbegin(), s.cend(), c) & 1;

        }

        return s.length() - odds + (odds > 0);

    }

};

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