409. Longest Palindrome

最新推荐文章于 2024-12-15 17:34:20 发布

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#Hash Table #Leetcode #Longest Palindrome

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本文探讨了如何从给定的包含大小写字母的字符串中构建最长的回文子串。通过统计每种字符出现的次数，寻找可以配对形成回文的字符，并考虑奇数次出现的字符作为回文中心的情况。

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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.

思路是统计每个字母出现的次数，然后累加次数除以2的商，也就是找相同的字母凑对儿，最后比较count和原字符串想不想等，不等说明有落单，可以加到回文字符串的最中间。实现方式不同，速度不同。采用hashmap比用数组直接存次数要慢。hashmap代码如下：

public class Solution {
    public int longestPalindrome(String s) {
        HashMap<Character, Integer> hm = new HashMap<Character, Integer>();
        char[] chars = s.toCharArray();
        for (int i = 0; i < chars.length; i ++) {
            if (hm.containsKey(chars[i])) {
                hm.put(chars[i], hm.get(chars[i]) + 1);
            } else {
                hm.put(chars[i], 1);
            }
        }
        int count = 0;
        Iterator iter = hm.entrySet().iterator();
        while (iter.hasNext()) {
            Map.Entry entry = (Map.Entry) iter.next();
            Integer value = (Integer)entry.getValue();
            count += value >> 1;
        }
        count = count << 1;
        if (count != chars.length) {
            count ++;
        }
        return count;
    }
}

数组直接存的代码如下：

public int longestPalindrome(String s) {
    int[] lowercase = new int[26];
    int[] uppercase = new int[26];
    int res = 0;
    for (int i = 0; i < s.length(); i++){
        char temp = s.charAt(i);
        if (temp >= 97) lowercase[temp-'a']++;
        else uppercase[temp-'A']++;
    }
    for (int i = 0; i < 26; i++){
        res+=(lowercase[i]/2)*2;
        res+=(uppercase[i]/2)*2;
    }
    return res == s.length() ? res : res+1;
        
}