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.

题意：

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