LeetCode387. 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.

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思路：一开始的想法就是用count()方法遍历，找到第一个count为1的字母，如下面这种方法，不过这样会超时。

class Solution:
    def firstUniqChar(self, s):
        """
        :type s: str
        :rtype: int
        """
        for i,char in enumerate(s):
            if s.count(char)==1:
                return i
        return -1

然后想到利用collections的Counter()方法，先统计个数，再对个数为1的进行位置比较，返回第一个个数为1字母的位置。

from collections import Counter
class Solution:
    def firstUniqChar(self, s):
        """
        :type s: str
        :rtype: int
        """
        min = float('inf')
        dic = Counter(s)

        for key,val in dic.items():
            if val==1:
                if min>s.index(key):
                    min = s.index(key)
        if min!=float('inf'):
            return min

        return -1

看了discuss，有更快的方法,不过要事先设好输入的letters。

class Solution:
    def firstUniqChar(self, s):
        """
        :type s: str
        :rtype: int
        """
        letters='abcdefghijklmnopqrstuvwxyz'

        index=[s.index(l) for l in letters if s.count(l) == 1]

        return min(index) if len(index) > 0 else -1