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.
Solution1：Hashmap(count) twopass
思路：
Time Complexity: O(N) Space Complexity: O(N)
Solution2：Array(count) twopass
思路：
Time Complexity: O(N) Space Complexity: O(N)
Solution1 Code：
class Solution {
public int firstUniqChar(String s) {
Map<Character, Integer> count_map = new HashMap<>();
for(int i = 0; i < s.length(); i++) {
count_map.put(s.charAt(i), count_map.getOrDefault(s.charAt(i), 0) + 1);
}
for(int i = 0; i < s.length(); i++) {
if(count_map.get(s.charAt(i)) == 1) {
return i;
}
}
return -1;
}
}
Solution2 Code：
public class Solution {
public int firstUniqChar(String s) {
int freq [] = new int[26];
for(int i = 0; i < s.length(); i ++)
freq [s.charAt(i) - 'a'] ++;
for(int i = 0; i < s.length(); i ++)
if(freq [s.charAt(i) - 'a'] == 1)
return i;
return -1;
}
}
