143. Reorder List Solution

Author: doh6077

reorder-list/doh6077.py

@@ -0,0 +1,43 @@
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, val=0, next=None):
+#         self.val = val
+#         self.next = next
+class Solution:
+    def reorderList(self, head: Optional[ListNode]) -> None:
+        """
+        Do not return anything, modify head in-place instead.
+        """
+        # Time Complexity: O(N)
+        # The order: 
+        # 0, n, 1, n -1, 2, n-3 ...
+        # first Idea
+        # need to save the index of the original head 
+        # Hashmap: iterate through the head until head it none, save index as key and head.val as value 
+        head_hash = {}
+        temp = head
+        index = 0 
+        length = 0 
+        # Save index and value in the hashmap
+        while temp is not None:
+            head_hash[index] = temp.val
+            temp = temp.next 
+            index += 1
+            length += 1
+        # reset index to 0, and use it to iterate through the head again
+        index = 0 
+        # to keep track of n-1, n-2, n-3 ...
+        count = 1
+        # Iterate through the head again and change the value based on the index
+        while head is not None: 
+            res = index % 2 
+            # if the index is even number 
+            if res == 0:
+                head.val = head_hash[index/2]
+            # n, n-1, n-2 when the index is odd 
+            else:
+                head.val = head_hash[length - count]
+                count += 1 
+            index += 1
+            head = head.next
+