387. First Unique Character in a String

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.

Solution1：Hashmap(count) twopass

思路：
Time Complexity: O(N) Space Complexity: O(N)

Solution2：Array(count) twopass

思路：
Time Complexity: O(N) Space Complexity: O(N)

Solution1 Code：

class Solution {
    public int firstUniqChar(String s) {
        Map<Character, Integer> count_map = new HashMap<>();
        for(int i = 0; i < s.length(); i++) {
            count_map.put(s.charAt(i), count_map.getOrDefault(s.charAt(i), 0) + 1);
        }
        
        for(int i = 0; i < s.length(); i++) {
            if(count_map.get(s.charAt(i)) == 1) {
                return i;
            }
        }
        return -1;
        
    }
}

Solution2 Code：

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;
    }
}