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
class Solution {
public int firstUniqChar(String s) {
if (s == null || s.length()) return -1;
Map<Character, Integer> map = new HashMap<>();
// build a map with char and frequency as <key, value> pair
for (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 = 1
for (int i = 0; i < s.length(); i++) {
if (map.get(s.charAt(i)) == 1) return i;
}
return -1;
}
}