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.
Solution
Straight solution, build a map of frequency,
build a map with [char:frequency] as pair
iterate through chars in string s, return index of the first char with frequency = 1, map.get(ch)==1
Code
classSolution {publicintfirstUniqChar(String s) {if (s ==null||s.length()) return-1;Map<Character,Integer> map =newHashMap<>();// build a map with char and frequency as <key, value> pairfor (char ch :s.toCharArray()) {map.put(ch,map.getOrDefault(ch,0) +1); }// iterate through chars in s, return the index of the first char with frequency = 1for (int i =0; i <s.length(); i++) {if (map.get(s.charAt(i)) ==1) return i; }return-1; }}