Lintcode627 - Longest Palindrome - easy

Posted 2020-12-27 jasminemzy

Description

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.

Assume the length of given string will not exceed 1010.

Example

Given s = "abccccdd" return 7

One longest palindrome that can be built is "dccaccd", whose length is 7.

 

算法：

问题转化为：计算字母匹配对数*2，若存在落单字母额外+1。（或者也可以转化为：总字符串长度-落单字母数，若存在落单字母额外+1）

实现用set。

 

细节：

1. 注意每次成对后计数器是+=2， 而不是++ （当然为那一点效率也可以++最后再*=2）

2. 注意回文串中间可以插一个字符这个特性。

 

自己实现：

public class Solution {
    /**
     * @param s: a string which consists of lowercase or uppercase letters
     * @return: the length of the longest palindromes that can be built
     */
    public int longestPalindrome(String s) {
        // write your code here
        
        int cnt = 0;
        if (s == null) {
            return cnt;
        }
        
        Set<Character> set = new HashSet<>();
        for (int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);
            if (set.contains(c)) {
                // not only plus 1
                cnt += 2;
                set.remove(c);
            } else {
                set.add(c);
            }
        }
        
        // !! you can insert one in the middle
        if (set.size() > 0) {
            cnt++;
        }
        return cnt;
    }
}

 

九章实现

public class Solution {
    /**
     * @param s a string which consists of lowercase or uppercase letters
     * @return the length of the longest palindromes that can be built
     */
    public int longestPalindrome(String s) {
        // Write your code here
        Set<Character> set = new HashSet<>();
        for (char c : s.toCharArray()) {
            if (set.contains(c)) set.remove(c);
            else set.add(c);
        }

        int remove = set.size();
        if (remove > 0)
            remove -= 1;
        return s.length() - remove;
    }
}