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.
解法一：cc150解法
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; } }
reference：
https://discuss.leetcode.com/topic/55148/java-7-lines-solution-29ms
解法二：hashmap
public class Solution { public int firstUniqChar(String s) { if(s.length() == 0) return -1; Map<Character, Integer> hm = new HashMap<>(); for(int i = 0; i < s.length(); i++){ if(!hm.containsKey(s.charAt(i))){ hm.put(s.charAt(i), i); }else{ hm.put(s.charAt(i), Integer.MAX_VALUE); } } return Collections.min(hm.values()) != Integer.MAX_VALUE ? Collections.min(hm.values()) : -1; } }
reference：
https://discuss.leetcode.com/topic/55607/simple-java-solution-using-hashmap-and-collections
First Unique Character in a String
原文：http://www.cnblogs.com/hygeia/p/5820385.html