387. First Unique Character in a String

题目摘要
给定一个字符串，返回第一个没有重复的字母的下标，如果不存在，返回-1。假设字符串中只含有小写字母。

解法
两个指针slow=0,fast=1;fast先走，并将经过的字母记录下来，直到c[fast] == c[slow],然后slow开始走，直到走到下一个没有记录过两次及以上的字母，然后fast赋为fast+1或slow+1中较大的那个。

public class Solution {    public int firstUniqChar(String s) {        if (s == null || s.length() == 0)             return -1;        if (s.length() == 1)            return 0;        int[] letters = new int[26];        for (int i = 0; i < 26; i++) {            letters[i] = 0;        }        char[] c = s.toCharArray();        int slow = 0, fast = 1;        letters[c[slow] - 'a'] = 1;        while (fast < c.length) {            while (c[fast] != c[slow]) {                letters[c[fast++] - 'a']++;                if (fast == c.length)                    return slow;            }            letters[c[fast] - 'a']++;            while (++slow < c.length) {                if (letters[c[slow] - 'a'] <= 1) {                    letters[c[slow] - 'a']++;                    break;                }            }            if (fast < slow)                fast = slow + 1;            else                fast++;        }        return slow == c.length ? -1 : slow;    }}

注意

*循环之前，现将第一个字母对应的数组位置加一

可问问题

原题
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.