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.

class Solution {
public:
    int firstUniqChar(string s) {
        int table[26] = {0};
        for (char ch : s)
            table[ch - 'a']++;
        
        for (int i = 0; i < s.size(); i++)
            if (table[s[i] - 'a'] == 1) return i;
            
        return -1;
    }
};

class Solution {
public:
    int firstUniqChar(string s) {
        unordered_map<char, bool> um;
        for (int i = 0; i < s.size(); i++) {
            if (!um[s[i]] && s.find(s[i], i + 1) == s.npos) return i;
            um[s[i]] = true;
        }
        return -1;
    }
};

class Solution {
public:
    int firstUniqChar(string s) {
        unordered_map<char, vector<int> > um;
        for (int i = 0; i < s.size(); i++)
            um[s[i]].push_back(i);
        
        int result = INT_MAX;
        for (auto item : um) {
            if (item.second.size() == 1) result = min(result, item.second[0]);
        }
        return result == INT_MAX ? -1 : result;
    }
};