409. Longest Palindrome

Very straightforward solution. Count the frequency for characters in the input string. If a character appears even times, it must be a part of palindrome. On the other hand, if a character appears odd times, then we want to add (odd times - 1). In the end, if there is odd time character, the return value should be added one.

1:  class Solution {  
2:  public:  
3:    int longestPalindrome(string s) {  
4:      unordered_map<char, int> mp;  
5:      for (c : s) mp[c]++;  
6:      int res = 0, odd = 0;  
7:      for (auto it : mp) {  
8:        if (it.second & 1) { res += it.second - 1, odd = 1; }  
9:        else res += it.second;  
10:      }  
11:      return res + odd;  
12:    }  
13:  };