Add the second solution for Valid Palindrome problem.

Author: KwonNayeon

valid-palindrome/KwonNayeon.py

@@ -3,17 +3,40 @@
 - 1 <= s.length <= 2 * 10^5
 - s consists only of printable ASCII characters.
 
-Time Complexity:
-- O(n)
-Space Complexity:
-- O(n)
+<Solution 1>
+Time Complexity: O(n)
+
+Space Complexity: O(n)
 """
-# Solution 1
 class Solution:
     def isPalindrome(self, s: str) -> bool:
         s = re.sub(r'[^a-zA-z0-9]', '', s).lower()
         if s == s[::-1]:
             return True
         return False
+"""
+<Solution 2>
+Time Complexity: O(n) 
+- 팰린드롬일 경우, 각 문자를 한 번씩 검사
+
+Space Complexity: O(1)
+- left, right 포인터 외에 추가 공간 사용 없음
+"""
+class Solution:
+    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
 
-# Solution 2
+        return True