5/27 每日一題(返回唯一值的索引)

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

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.

 

class Solution:
    def firstUniqChar(self, s: str) -> int:
        s_c=Counter(s) #建立一個key=字元,value=出現次數的字典
        for i,j in s_c.items():
            if j==1: #如果有唯一值,返回該元素的索引值
                return s.index(i)
        return -1 #迴圈結束表示不存在唯一值

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參考解答

1.計算第一次出現並打破循環
class Solution:
    def firstUniqChar(self, s: str) -> int:
        for i,c in enumerate(s):
            if s.count(c)==1:
                return i
                break
        return -1
    #please upvote me it would encourage me alot
2. 使用哈希表和集合
class Solution:
    def firstUniqChar(self, s: str) -> int:
        dic={}
        seen=set()
        for ind,let in enumerate(s):
            if let not in seen:
                dic[let]=ind
                seen.add(let)
            elif let in dic:
                del dic[let]
        return min(dic.values()) if dic else -1

    //please upvote me it would encourage me alot