234 Update

Author: tech-cow

linkedlist/palindrome.py

@@ -0,0 +1,40 @@
+# 234. Palindrome Linked List
+
+# Given a singly linked list, determine if it is a palindrome.
+# Time: O(n)
+# Space: O(1)
+
+# 思路：
+# 设置快慢pointer，当快的pointer指向最后的时候，慢的正好是中间
+# 以中间的Node为起始，开始reverse后半边的链表
+# 从Head和Tail开始往中间循环，比对
+
+class Solution(object):
+    def isPalindrome(self, head):
+        """
+        :type head: ListNode
+        :rtype: bool
+        """
+        slow = fast = head
+
+        while fast and fast.next:
+            fast = fast.next.next
+            slow = slow.next
+
+        prev = None
+
+        #Reverse here
+        while slow:
+            nxt = slow.next
+            slow.next = prev
+            prev = slow
+            slow = nxt
+
+        tail = prev
+
+        while head and tail:
+            if head.val != tail.val:
+                return False
+            head = head.next
+            tail = tail.next
+        return True