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Formatted question description: https://leetcode.ca/all/409.html

# 409. Longest Palindrome (Easy)

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.


Related Topics:
Hash Table

Similar Questions:

## Solution 1.

• class Solution {
public int longestPalindrome(String s) {
char[] array = s.toCharArray();
int[] upperCounts = new int[26];
int[] lowerCounts = new int[26];
for (char c : array) {
if (c <= 'Z')
upperCounts[c - 'A']++;
else
lowerCounts[c - 'a']++;
}
boolean oddFlag = false;
int count = 0;
for (int i = 0; i < 26; i++) {
int upperCount = upperCounts[i], lowerCount = lowerCounts[i];
count += upperCount / 2 * 2;
count += lowerCount / 2 * 2;
if (upperCount % 2 != 0 || lowerCount % 2 != 0)
oddFlag = true;
}
if (oddFlag)
count++;
return count;
}
}

############

class Solution {
public int longestPalindrome(String s) {
int[] cnt = new int[128];
for (int i = 0; i < s.length(); ++i) {
++cnt[s.charAt(i)];
}
int ans = 0;
for (int v : cnt) {
ans += v - (v & 1);
if (ans % 2 == 0 && v % 2 == 1) {
++ans;
}
}
return ans;
}
}

• // OJ: https://leetcode.com/problems/longest-palindrome/
// Time: O(N)
// Space: O(1) because there are at most 52 unique characters
class Solution {
public:
int longestPalindrome(string s) {
unordered_map<int, int> cnt;
for (char c : s) cnt[c]++;
int ans = 0, odd = 0;
for (auto &p : cnt) {
if (p.second % 2) {
odd = 1;
ans += p.second - 1;
} else ans += p.second;
}
return ans + odd;
}
};

• class Solution:
def longestPalindrome(self, s: str) -> int:
n = len(s)
counter = Counter(s)
odd_cnt = sum(e % 2 for e in counter.values())
return n if odd_cnt == 0 else n - odd_cnt + 1

############

class Solution(object):
def longestPalindrome(self, s):
"""
:type s: str
:rtype: int
"""
maxLen = 0
single = False
d = {}
for c in s:
d[c] = d.get(c, 0) + 1

for key in d:
if d[key] >= 2:
count = d[key]
left = d[key] % 2
d[key] = left
maxLen += count - left
if not single:
if d[key] == 1:
maxLen += 1
single = True
return maxLen


• func longestPalindrome(s string) (ans int) {
cnt := [128]int{}
for _, c := range s {
cnt[c]++
}
for _, v := range cnt {
ans += v - (v & 1)
if ans&1 == 0 && v&1 == 1 {
ans++
}
}
return
}

• function longestPalindrome(s: string): number {
let n = s.length;
let ans = 0;
let record = new Array(128).fill(0);
for (let i = 0; i < n; i++) {
record[s.charCodeAt(i)]++;
}
for (let i = 65; i < 128; i++) {
let count = record[i];
ans += count % 2 == 0 ? count : count - 1;
}
return ans < s.length ? ans + 1 : ans;
}


• use std::collections::HashMap;

impl Solution {
pub fn longest_palindrome(s: String) -> i32 {
let mut map: HashMap<char, i32> = HashMap::new();
for c in s.chars() {
map.insert(c, map.get(&c).unwrap_or(&0) + 1);
}
let mut has_odd = false;
let mut res = 0;
for v in map.values() {
res += v;
if v % 2 == 1 {
has_odd = true;
res -= 1;
}
}
res + if has_odd { 1 } else { 0 }
}
}