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.

Brute force solution, traverse string s 2 times. First time, store counts of every character into the hash table,
second time, find the first character that appears only once.
 

if the string is extremely long, we wouldn‘t want to traverse it twice,
 so instead only storing just counts of a char, we also store the index, and then traverse the hash table.

class Solution {
public:
    int firstUniqChar(string s) {
        unordered_map<char, int> m;
        for(auto &c : s) {
            m[c]
 
++;
        }
        for(int i = 0; i < s.size(); i++) {
            if(m[s[i]] == 1) return i;
        }
        return -1;
    }
};

class Solution {
public:
    int firstUniqChar(string s) {
        unordered_map<char, pair<int, int>> m;
        int idx = s.size();
        for(int
 
 i = 0; i < s.size(); i++) {
            m[s[i]].first++;
            m[s[i]].second = i;
        }
        for(auto &p : m) {
            if(p.second.first == 1) idx = min(idx, p.second.second);
        }
        return idx == s.size() ? -1 : idx;
    }
};

參考：
https://discuss.leetcode.com/topic/55082/c-2-solutions

[leetcode-387-First Unique Character in a String]