Palindrome Linked List/ Valid Anagram

Author: gavinfish

234 Palindrome Linked List.py

@@ -0,0 +1,49 @@
+'''
+Given a singly linked list, determine if it is a palindrome.
+
+Follow up:
+Could you do it in O(n) time and O(1) space?
+'''
+
+# Definition for singly-linked list.
+class ListNode(object):
+    def __init__(self, x):
+        self.val = x
+        self.next = None
+
+
+class Solution(object):
+    def isPalindrome(self, head):
+        """
+        :type head: ListNode
+        :rtype: bool
+        """
+        # split the list into two parts, we can reverse the first part at meantime
+        fast = slow = head
+        reverse = None
+        while fast and fast.next:
+            fast = fast.next.next
+            reverse, reverse.next, slow = slow, reverse, slow.next
+        # if the total number is odd, skip the centre point
+        if fast:
+            slow = slow.next
+        # compare the reversed first part and normal second part
+        while reverse:
+            if reverse.val != slow.val:
+                return False
+            else:
+                reverse = reverse.next
+                slow = slow.next
+        return True
+
+
+if __name__ == "__main__":
+    n1 = ListNode(1)
+    n2 = ListNode(2)
+    n3 = ListNode(1)
+    n1.next = n2
+    n2.next = n3
+    assert Solution().isPalindrome(n1) == True
+    assert Solution().isPalindrome(n2) == False
+    n1.next = n3
+    assert Solution().isPalindrome(n1) == True

242 Valid Anagram.py

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+'''
+Given two strings s and t, write a function to determine if t is an anagram of s.
+
+For example,
+s = "anagram", t = "nagaram", return true.
+s = "rat", t = "car", return false.
+
+Note:
+You may assume the string contains only lowercase alphabets.
+
+Follow up:
+What if the inputs contain unicode characters? How would you adapt your solution to such case?
+'''
+
+class Solution(object):
+    def isAnagram(self, s, t):
+        """
+        :type s: str
+        :type t: str
+        :rtype: bool
+        """
+        m = {}
+        if len(s) != len(t):
+            return False
+        for c in s:
+            m[c] = m.get(c, 0) + 1
+        for c in t:
+            if c not in m or m[c] == 0:
+                return False
+            else:
+                m[c] -= 1
+        return True
+
+
+if __name__ == "__main__":
+    s, t = "anagram", "nagaram"
+    assert Solution().isAnagram(s, t) == True
+    s, t = 'rat', 'car'
+    assert Solution().isAnagram(s, t) == False