Solve valid-palindrome in two-way

Author: OstenHun

valid-palindrome/OstenHun.cpp

@@ -0,0 +1,84 @@
+/*
+    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.
+
+    Constraints:
+
+    1 <= s.length <= 2 * 105
+    s consists only of printable ASCII characters.
+*/
+
+#include <cctype>
+#include <string>
+#include <vector>
+using namespace std;
+
+#pragma region ExtraSpaceIdea
+// 새로운 배열을 만들어서 문자열을 정리한 후에 팰린드롬 판별
+// 시간 복잡도 : O(n)
+// 공간 복잡도 : O(n)
+// n은 문자열의 길이일 것이다.
+namespace extra_space_idea {
+
+class Solution {
+public:
+    bool isPalindrome(string s) {
+        vector<char> str;
+        str.reserve(s.size());
+
+        size_t len = s.length();
+        for (size_t i = 0; i < len; i++) {
+            if (isalnum(s[i])) {
+                str.push_back(tolower(s[i]));
+            }
+        }
+
+        size_t vec_len = str.size();
+        for (size_t i = 0; i < (vec_len + 1) / 2; i++) {
+            if (str[i] != str[vec_len - 1 - i]) return false;
+        }
+
+        return true;
+    }
+};
+
+}  // namespace extra_space_idea
+#pragma endregion
+
+#pragma region FinalSolution
+// 투 포인터를 이용해 추가 배열 없이 구현
+// 시간 복잡도 : O(n)
+// 공간 복잡도 : O(1)
+class Solution {
+public:
+    bool isPalindrome(string s) {
+        int left = 0;
+        int right = s.length() - 1;
+
+        while(left < right) {
+            if (!isalnum(s[left]))
+                left++;
+            else if (!isalnum(s[right]))
+                right--;
+            else {
+                if (tolower(s[left]) != tolower(s[right]))
+                    return false;
+                left++;
+                right--;
+            }
+        }
+
+        return true;
+    }
+};
+#pragma endregion