LeetCode(五)387. First Unique Character in a String&409. Longest Palindrome-优快云博客

本文针对两道经典字符串算法题目提供了解决方案，包括查找字符串中第一个唯一字符的位置及构建最长回文子串的方法。通过不同数据结构的应用，如数组、哈希集等，展示多种解题思路。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

387. First Unique Character in a String

题目要求如下

Given a string, find the first non-repeating character in it and
return it’s index. If it doesn’t exist, return -1.

Examples:

s = “leetcode” return 0.

s = “loveleetcode”, return 2.

Note: You may assume the string contain
only lowercase letters.

题目解法如下

public class Solution {
    public int firstUniqChar(String s) {
        int freq[]=new int[26];
        for(int i=0;i<s.length();i++){
            freq[s.charAt(i)-'a']++;
        }
        for(int i=0;i<s.length();i++){
            if(freq[s.charAt(i)-'a']==1)
                return i;
        }
        return -1; 
    }
}

409. Longest Palindrome

题目要求如下

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.

要求最长的回文数，大小写敏感
参考上一题，我的解法如下

public class Solution {
    public int longestPalindrome(String s) {
        int[]alphabet=new int[58];
        for(int i=0;i<s.length();i++){
            alphabet[s.charAt(i)-'A']+=1;
        }
        int count=0;
        for(int j=0;j<58;j++){
            if(alphabet[j]>1){
                if(alphabet[j]%2==0){
                    count+=alphabet[j];
                }
                else{
                    count+=alphabet[j]-1;
                }
            }
        }
        return count==s.length()?count:count+1;
    }
}

也可以使用两个int[26]集合

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;

}

使用HashSet的方法

public int longestPalindrome(String s) {
        if(s==null || s.length()==0) return 0;
        HashSet<Character> hs = new HashSet<Character>();
        int count = 0;
        for(int i=0; i<s.length(); i++){
            if(hs.contains(s.charAt(i))){
                hs.remove(s.charAt(i));
                count++;
            }else{
                hs.add(s.charAt(i));
            }
        }
        if(!hs.isEmpty()) return count*2+1;
        return count*2;
}