problem

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.

• 在字符串中寻找第一个不重复的字母，返回位置（没有则返回-1）

solution

1. 求一个重复字母的集合。
2. 寻找第一个不在集合的元素

class Solution(object):    def firstUniqChar(self, s):        """        :type s: str        :rtype: int        """        double = []                for x in set(s):            if s.count(x)>1:                double.append(x)                        for x,y in enumerate(s):            if y not in double:                return x         return -1

discuss solution

• 建一个包含26个字母数值的list；第一遍循环，每发现一次某字母，对其做一个标示；第二次循环，发现字母标示为（出现1次）
• 根据字母index（first one == last one）（java特性？ s.indexOf(a[i])==s.lastIndexOf(a[i])）

str.rfind(sub[, start[, end]])Return the highest index in the string where substring sub is found, such that sub is contained within s[start:end]. Optional arguments start and end are interpreted as in slice notation. Return -1 on failure.str.rindex(sub[, start[, end]])Like rfind() but raises ValueError when the substring sub is not found.

if s.find(x)==s.rfind(x):    return s.find(x)