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.

自己的解法

开始觉的遍历 char[] 然后判断在 String 的首位置和末位置，若一样就返回索引。呃，感觉有点偷鸡

public int firstUniqChar(String s) {
  for (int i = 0; i < s.length(); i++) {
    if (s.indexOf(s.charAt(i)) == s.lastIndexOf(s.charAt(i))) {
      return i;
    }
  }
  return -1;
}

查了下别人的，又是 new int[26]

public int firstUniqChar(String s) {
  int[] count = new int[26];
  for (int i = 0; i < s.length(); i++) {
    count[s.charAt(i) - 'a']++;
  }
  for (int i = 0; i < s.length(); i++) {
    if (count[s.charAt(i) - 'a'] == 1) {
      return i;
    }
  }
  return -1;
}

但是在运行时间上，下面的方法更快