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

时间复杂度：o(N)