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LeetCode First Unique Character in a String 387

字数统计: 215阅读时长: 2 min
2016/08/23 Share

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

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  1. 1. 自己的解法