leetcode-387. First Unique Character in a String
387. First Unique Character in a String
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.
找出字符串 s
中第一个不重复的字符,并返回他的序号
解法:
一开始就想着用map来记录每次字符出现的次数,然后找到出现一次的字符就返回序号就可以了
但是提交过后发现用时太久了,用了 111ms
class Solution { public int firstUniqChar(String s) { if(s.length()==0) return -1; int solu = -1; Map<Character,Integer> map = new HashMap<>(); for (int i = 0; i < s.length(); i++) { if(map.get(s.charAt(i))!= null){ map.put(s.charAt(i),map.get(s.charAt(i))+1); }else{ map.put(s.charAt(i),1); } } for (int i = 0; i < s.length(); i++) { if(map.get(s.charAt(i))==1) { solu = i; break; } } return solu; } }//111ms
看了diss中的解法,发现用数组就会快很多很多,,思路还是一致的。
class Solution { public int firstUniqChar(String s) { int[] arr = new int[27]; for (int i = 0;i < s.length() ;i++ ) arr[s.charAt(i) - 'a'] ++; for (int i = 0;i< s.length() ;i++ ) if(arr[s.charAt(i) - 'a'] == 1) return i; return -1; } }//28ms