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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class Solution(object):    def firstUniqChar(self, s):        """        :type s: str        :rtype: int        """        cnt = collections.Counter(s)        for i,c in enumerate(s):            if cnt[c] == 1:                return i        return -1

java 里还可以，直接一次扫描s搞定：

class Solution {    public int firstUniqChar(String s) {        Map
     
       map = new LinkedHashMap<>();        Set
      
        set = new HashSet<>();        for (int i = 0; i < s.length(); i++) {            if (set.contains(s.charAt(i))) {                if (map.get(s.charAt(i)) != null) {                    map.remove(s.charAt(i));                }            } else {                map.put(s.charAt(i), i);                set.add(s.charAt(i));            }        }        return map.size() == 0 ? -1 : map.entrySet().iterator().next().getValue();    }}
      
     

因为map里entryset保存了add顺序？？？