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387. First Unique Character in a String ​

题目 ​

Given a string s, find the first non-repeating character in it and return its index. If it does not exist, return -1.

 

Example 1:

Input: s = "leetcode"

Output: 0

Explanation:

The character 'l' at index 0 is the first character that does not occur at any other index.

Example 2:

Input: s = "loveleetcode"

Output: 2

Example 3:

Input: s = "aabb"

Output: -1

 

Constraints:

  • 1 <= s.length <= 105
  • s consists of only lowercase English letters.
Related Topics
  • 队列
  • 哈希表
  • 字符串
  • 计数

  • 👍 776
  • 👎 0

    • 思路 ​

    解法 ​

    py
    # leetcode submit region begin(Prohibit modification and deletion)
from collections import Counter, defaultdict


class Solution:
    def firstUniqChar(self, s: str) -> int:
        locations = {}
        dict = defaultdict(int)
        for index, value in enumerate(s):
            locations[value] = index
            dict[value] += 1

        for key, value in dict.items():
            if value == 1:
                return locations[key]

        return -1

        
# leetcode submit region end(Prohibit modification and deletion)

    复杂度分析 ​

    • 时间复杂度 O(N)
      • 遍历 S 所需要的时间 len(s)
    • 空间复杂度 O(N)
      • 创建 location hashmap 空间为 N = len(s)