LeetCode_ String_ Simple_ 387. first unique character in string

1. subject

Given a string s , find Its first non repeating character , And return its index . If it doesn't exist , Then return to -1 .

Example 1：
Input : s = “leetcode”
Output : 0

Example 2：
Input : s = “loveleetcode”
Output : 2

Example 3：
Input : s = “aabb”
Output : -1

Tips ：
1 <= s.length <= 10⁵
s Contains only lowercase letters

source ： Power button （LeetCode）
link ：https://leetcode.cn/problems/first-unique-character-in-a-string

2. Ideas

（1） Hashtable

（2） Array count

3. Code implementation （Java）

// Ideas 1———— Hashtable 
class Solution {
    
    public int firstUniqChar(String s) {
    
        //hashmap Used to store strings  s  Number of occurrences of each character in 
        Map<Character, Integer> hashmap = new HashMap<>();
        for (int i = 0; i < s.length(); i++) {
    
            char c = s.charAt(i);
            hashmap.put(c, hashmap.getOrDefault(c, 0) + 1);
        }
        // find  hashmap  The number of occurrences in is  1  The characters of , And return its subscript 
        for (int i = 0; i < s.length(); i++) {
    
            if (hashmap.get(s.charAt(i)) == 1) {
    
                return i;
            }
        }
        // There are no unique characters , return  -1
        return -1;
    }
}

// Ideas 2———— Array count 
class Solution {
    
    public int firstUniqChar(String s) {
    
        /*  character string  s  Contains only lowercase letters , Therefore, the service length is  26  Array of  freq  To store each character in  s  Is the number of times  freq[0] Storage  a  Number of occurrences ,freq[1] Storage  b  Number of occurrences , And so on . */
        int[] freq = new int[26];
        int length = s.length();
        for (int i = 0; i < length; i++) {
    
            freq[s.charAt(i) - 'a']++;
        }
        for (int i = 0; i < length; i++) {
    
            // If you find a character that appears only once , Just return its subscript directly 
            if (freq[s.charAt(i) - 'a'] == 1) {
    
                return i;
            }
        }
        // There are no unique characters , return  -1
        return -1;
    }
}