387. First Unique Character in a String

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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.

我的做法：

class Solution {public:    int firstUniqChar(string s) {     map<char, int> maps;    char target;    int min=-1;    for (char a : s)    {        maps[a]++;    }    for (map<char, int>::iterator iter = maps.begin(); iter != maps.end(); iter++)    {        if (iter->second == 1)        {            target = iter->first;            if (min == -1)                min = s.find(target);            if (min > s.find(target))            {                min = s.find(target);            }        }    }    return min;    }};

太慢了，太冗余，后来发现别人的做法跟我刚开始的想法相似，不过他解决了查询过程中，查的不是按照字母表先后顺序，而是按照字母在字符串中顺序的问题。

别人的做法：

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

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