LeetCode 125 Valid Palindrome

LeetCode

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 <= 2 10⁵*s` consists only of printable ASCII characters.

题目大意：

求含非字母数字的字符串是否回文，字符串含空格，冒号等. Easy题

双指针解题思路(推荐)：

回文首先考虑用相向双指针

解题步骤：

N/A

注意事项：

比较时，要转换成小写
外循环left < right条件要复制到内循环中

Python代码：

def isPalindrome(self, s: str) -> bool:
	left, right = 0, len(s) - 1
	while left < right:
		while left < right and not s[left].isalnum():
			left += 1
		while left < right and not s[right].isalnum():
			right -= 1
		if s[left].lower() != s[right].lower():
			return False
		left += 1
		right -= 1
	return True

算法分析：

时间复杂度为O(n)，空间复杂度O(1)

reverse法算法II解题思路：

reverse字符串比较

注意事项：

比较时，要转换成小写

Python代码：

def isPalindrome2(self, s: str) -> bool:
	res = ''
	for char in s:
		if char.isalpha() or char.isdigit():
			res += char.lower()
	return True if res == res[::-1] else False

算法分析：

时间复杂度为O(n)，空间复杂度O(n)