前言

大家好，我是毛毛。ヾ(´∀ ˋ)ﾉ
來到Two Pointers的部分~
前面Array & Hashing有一題破不了，Encode and Decode Strings要訂閱才能寫QQ，於是Pass~XD

那就開始今天的解題吧~

Question

A phrase is a palindrome if, after converting all uppercase letters into lowercase letters and removing all non-alphanumeric characters, it reads the same forward and backward. Alphanumeric characters include letters and numbers.

Given a string s, return true if it is a palindrome, or false otherwise.

Example 1:

Input: s = "A man, a plan, a canal: Panama"
Output: true
Explanation: "amanaplanacanalpanama" is a palindrome.

Example 2:

Input: s = "race a car"
Output: false
Explanation: "raceacar" is not a palindrome.

Example 3:

Input: s = " "
Output: true
Explanation: s is an empty string "" after removing non-alphanumeric characters.
Since an empty string reads the same forward and backward, it is a palindrome.

Constraints:

• 1 <= s.length <= 2 * 10^5
• s consists only of printable ASCII characters.

Think

簡單的說，就是要判斷字串是不是回文的格式。

只是這題的測資有很多特殊符號需要移除掉QQ，測了好多次才找到全部的符號...

Code

class Solution:
    def isPalindrome(self, s: str) -> bool:
        s = s.strip()
        for character in ['\\', '`', '*', '_', '{',\
                          '}', '[', ']', '(', ')', \
                          '>', '#', '+', '-', '.', 
                          '!', '$', '\'', ' ', ',',\
                          ':', '%', '^', '<', '&',\
                          '=', '?', '{', '}', '|',\
                          '/', '~', '@', '\"', ';']:
            if character in s:
                s = s.replace(character,"")
        s = s.lower()
        # print(s)
        
        low = 0
        high = len(s)-1
        
        while(low<=high):
            if s[low] != s[high]:
                return False
            low += 1
            high -= 1
        
        return True

Submission

今天就到這邊啦~
大家明天見