README_EN.md

409. Longest Palindrome

中文文档

Description

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

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

 

Example 1:

Input: s = "abccccdd"
Output: 7
Explanation:
One longest palindrome that can be built is "dccaccd", whose length is 7.

Example 2:

Input: s = "a"
Output: 1

Example 3:

Input: s = "bb"
Output: 2

 

Constraints:

• 1 <= s.length <= 2000
• s consists of lowercase and/or uppercase English letters only.

Solutions

Python3

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

Java

class Solution {
    public int longestPalindrome(String s) {
        int[] counter = new int[128];
        for (char c : s.toCharArray()) {
            ++counter[c];
        }
        int oddCnt = 0;
        for (int e : counter) {
            oddCnt += (e % 2);
        }
        int n = s.length();
        return oddCnt == 0 ? n : n - oddCnt + 1;
    }
}

TypeScript

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;
};

C++

class Solution {
public:
    int longestPalindrome(string s) {
        vector<int> counter(128);
        for (char c : s) ++counter[c];
        int oddCnt = 0;
        for (int e : counter) oddCnt += e % 2;
        int n = s.size();
        return oddCnt == 0 ? n : n - oddCnt + 1;
    }
};

Go

func longestPalindrome(s string) int {
	counter := make([]int, 128)
	for _, c := range s {
		counter[c]++
	}
	oddCnt := 0
	for _, e := range counter {
		oddCnt += e % 2
	}
	n := len(s)
	if oddCnt == 0 {
		return n
	}
	return n - oddCnt + 1
}

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