125. Valid Palindrome

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.

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Mysolution:

• 这里可以考虑使用双指针的解法，一个left从左到右，一个right从右到左
• 其中对每个指针，我们想跳过中间非字母数字的部分
• 我们可以通过ASCII码来判断是不是数字或字母

class Solution:
    def isPalindrome(self, s: str) -> bool:
        left, right = 0, len(s) - 1

        while left < right:
            l, r = str.lower(s[left]),str.lower(s[right])
            if not (ord("a") <= ord(l) <= ord("z") or ord("0") <= ord(l) <= ord("9")):
                left += 1
            elif not (ord("a") <= ord(r) <= ord("z") or ord("0") <= ord(r) <= ord("9")):
                right -= 1
            elif l == r:
                left += 1
                right -= 1
            else:
                return False
        
        return True

Problem:

• ASCII的比较写起来麻烦且被重复复用，可以考虑写一个helper function方便复用
• 这里用了left，right又用了l，r，命名不清晰
• 可以先判断是不是alphanumeric，是的话再转大小写

class Solution:
		def isPalindrome(self, s: str) -> bool:
				l, r = 0, len(s) - 1
				while l < r:
						while l < r and not self.isAlphanumeric(s[l]):
								l += 1
						while r > l and not self.isAlphanumeric(s[r]):
								r -= 1
						if str.lower(s[l]) != str.lower(s[r]):
								return False
						l += 1
						r -= 1
				return True 
				
		def isAlphanumeric(self, c: chr) -> bool:
				return (ord("a") < ord(c) < ord("z") or
						ord("A") < ord(c) < ord("Z") or
						ord("0") < ord(c) < ord("9"))

392. Is Subsequence

Given two strings s and t, return true if s is a subsequence of t, or false otherwise.

A subsequence of a string is a new string that is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (i.e., "ace" is a subsequence of "abcde" while "aec" is not).

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Mysolution:

• 双指针写法，从左到右遍历t，如果遍历到的元素刚好和s里目前处理的元素的一致，则将s中处理的元素右移一位