力扣题目汇总（独特的电子邮件地址,二进制求和,反转字符串中的元音字母）

输入：["test.email+alex@leetcode.com","test.e.mail+bob.cathy@leetcode.com","testemail+david@lee.tcode.com"]
输出：2
解释：实际收到邮件的是 "testemail@leetcode.com" 和 "testemail@lee.tcode.com"。

#根据题意走删除没必要的东西,组成新的列表,最后用集合去重返回长度

class Solution:
    def numUniqueEmails(self, emails: List[str]) -> int:
        res = set()      
        for email in emails:
            head, tail = email.split('@')
            head = head.split('+')[0]
            head = head.replace('.', '')
            res.add(head+"@"+tail)
        return len(res)

输入: a = "11", b = "1"
输出: "100"

输入: a = "1010", b = "1011"
输出: "10101"

先变成二进制进行加再把返回值中的0b去掉

class Solution:
    def addBinary(self, a: str, b: str) -> str:
        return bin(int(a,2)+int(b,2))[2:]

输入: "hello"
输出: "holle"

输入: "leetcode"
输出: "leotcede"

关键在你与有大写和小写,而且题意这样出他的元音字母肯定是偶数个

class Solution:
    def reverseVowels(self, s: str) -> str:
        l1 = ['a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U']
        i, j = 0, len(s)-1
        s = list(s)
        while i < j:
            if s[i] in l1 and s[j] in l1:
                s[i], s[j] = s[j], s[i]
                i += 1
                j -= 1
            if s[i] not in l1 or s[j] not in l1:
                if s[j] not in l1:
                    j -= 1
                if s[i] not in l1:
                    i += 1
        return ''.join(s)