題目：

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.

解釋： 返回字串中第一個僅出現一次的字母在陣列中的index。第一反應是遍歷字串，暴力count。 python程式碼：

from collections import Counter
class Solution(object):
    def firstUniqChar(self, s):
        """
        :type s: str
        :rtype: int
        """
        _dict=Counter(s)
        for i in xrange(len(s)):
            if _dict[s[i]]==1:
                return i  
        return -1

python中，有find()

from collections import Counter
class Solution(object):
    def firstUniqChar(self, s):
        """
        :type s: str
        :rtype: int
        """
        alphas="qwertyuiopasdfghjklzxcvbnm";
        min_index=
 
len(s)
        for letter in alphas:
            index1=s.find(letter)
            index2=s.rfind(letter)
            if index1!=-1 and index1==index2:
                min_index=min(min_index,index1)
        return min_index if min_index!=len(s) else -1

c++程式碼：

class Solution {
public:
    int firstUniqChar(string s) {
        string alphas="qwertyuiopasdfghjklzxcvbnm";
        unsigned long int min_index=s.size();
        for (auto letter:alphas)
        {   auto index1=s.find(letter);
            auto index2=s.rfind(letter);
            //cout<<letter<<" "<<index1<<" "<<index2<<endl;
            if (index1!=string::npos && index1==index2)
            {
                
                min_index=min(min_index,index1);
                //cout<<letter<<" "<<min_index<<endl;
                
            }
                
        }
        return min_index==s.size()?-1:min_index;
    }
};

總結： string find()系列函式，返回型別是size_type 實際上是unsigned long int，注意不是int，也注意不要把size_type 強制轉換為int。