solve : #220 (valid-palindrome) with Java

Author: kdh-92

valid-palindrome/kdh-92.java

@@ -0,0 +1,65 @@
+/**
+ * Constraints
+ * - 1 <= s.length <= 2 * 10^5
+ * - s consists only of printable ASCII characters.
+ *
+ * Output
+ * - true : 좌우 동일
+ * - false : 좌우 비동일
+ */
+
+
+class Solution {
+    public boolean isPalindrome(String s) {
+
+        // 해결법 1 (투포인터)
+        // 시간복잡도: O(N), 공간 복잡도 : O(1)
+        // 2ms & Beats 99.05%
+        int left = 0, right = s.length() - 1;
+
+        while (left < right) {
+            // (1)
+            if (!Character.isLetterOrDigit(s.charAt(left))) {
+                left++;
+            }
+            else if (!Character.isLetterOrDigit(s.charAt(right))) {
+                right--;
+            }
+            else {
+                if (Character.toLowerCase(s.charAt(left)) != Character.toLowerCase(s.charAt(right))) {
+                    return false;
+                }
+
+                left++;
+                right--;
+            }
+
+
+            // (2)
+            while (left < right && !Character.isLetterOrDigit(s.charAt(left))) {
+                left++;
+            }
+            while (left < right && !Character.isLetterOrDigit(s.charAt(right))) {
+                right--;
+            }
+            if (Character.toLowerCase(s.charAt(left)) != Character.toLowerCase(s.charAt(right))) {
+                return false;
+            }
+            left++;
+            right--;
+        }
+
+        return true;
+
+        // 해결법 2 (Stream API)
+        // 시간 복잡도 : O(N), 공간 복잡도 : O(N)
+        // 133ms & Beats 5.58%
+        String filtered = s.chars()
+                .filter(Character::isLetterOrDigit)
+                .mapToObj(c -> String.valueOf((char) Character.toLowerCase(c)))
+                .reduce("", String::concat);
+
+        return IntStream.range(0, filtered.length() / 2)
+                .allMatch(i -> filtered.charAt(i) == filtered.charAt(filtered.length() - 1 - i));
+    }
+}