Longest Palindrome

最新推荐文章于 2020-11-19 23:10:06 发布

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最新推荐文章于 2020-11-19 23:10:06 发布

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分类专栏： leetcode 文章标签： 2d string class

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本文介绍了一个算法问题：如何从给定字符串中构建最长的回文串。提供了Java和Python两种实现方式，通过统计字符出现次数来计算最长可能长度。

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.

javaclass Solution {
    public int longestPalindrome(String s) {
        if (s == null || s.length() == 0) {
            return 0;
        }
        int len = 0;
        int count = 0;
        int[] arr = new int[256];
        char[] c = s.toCharArray();
        for (int i = 0; i < c.length; i++) {
            arr[c[i]]++;
        }
        for (int i = 0; i < 256; i++) {
            if (arr[i] % 2 == 0) {
                len += arr[i];
            } else {
                count++;
                len += (arr[i] - 1);
            }
        }
        if (count == 0) {
            return len;
        } else {
            return len + 1;
        }
    }
}
python

class Solution(object):
    def longestPalindrome(self, s):
        """
        :type s: str
        :rtype: int
        """
        if s is None or len(s) == 0:
            return 0
        lenx, count = 0, 0
        arr = [0] * 256
        for c in s:
            arr[ord(c)] += 1
        for i in range(256):
            if arr[i] % 2 == 0:
                lenx += arr[i]
            else:
                lenx += arr[i] - 1
                count += 1
        if count == 0:
            return lenx
        else:
            return lenx + 1