String - 58. Length of Last Word

58. Length of Last Word

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.

思路：

找出第一个不重复出现的字母，这题最容易想到的是用一个map记录下每个字母出现的次数，然后根据map去寻找只出现一次的字母的索引。这样做的时间消耗太多，采用另一种思路：题目中的字母默认只会有小写字母，那就可以以a - z来遍历字符串，如果第一次出现的索引下标和最后一次相同，就是结果，然后记录这其中下标最小的。

代码：

java：

class Solution {

    /*public int firstUniqChar(String s) {
        if (s == null || s.length() == 0) return -1;
        
        HashMap<Character, Integer> count = new HashMap<Character, Integer>();
        for (int i = 0; i <s.length(); i++) {
            char c =  s.charAt(i);
            count.put(c, count.getOrDefault(c, 0) + 1);
        }
        
        for (int i = 0; i <s.length(); i++) {
            if (count.get(s.charAt(i)) == 1) return i;
        }
        
        return -1;
    }*/
    
    public int firstUniqChar(String s) {
        if (s == null || s.length() == 0) return -1;
        
        int res = Integer.MAX_VALUE;
        for (char c = 'a'; c <= 'z'; c++) {
            if (s.indexOf(c) == -1) continue;
            
            if (s.indexOf(c) == s.lastIndexOf(c)) 
                res = Math.min(res, s.indexOf(c));
        }
        
        return res == Integer.MAX_VALUE ? -1 : res;
    }
}