Create 160_intersection_of_two_linked_lists.md

Author: Joseph Luce

leetcode/easy/160_intersection_of_two_linked_lists.md

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+# 160. Intersection of Two Linked Lists
+
+## Solution with dictionary
+- Runtime: O(N)
+- Space: O(1)
+- N = Number of nodes in linked list
+
+If you first count the amount of nodes in both linked list, you can then figure out the difference in the number of nodes between the two.
+Using that, you can then move the starting pointer of the longest linked list to the position where you can iterate both linked lists at the same time to find the intersection node.
+
+```
+class Solution(object):
+    def getIntersectionNode(self, headA, headB):
+        """
+        :type head1, head1: ListNode
+        :rtype: ListNode
+        """
+        def get_n_nodes(root):
+            if root is None:
+                return 0
+            n_nodes = 1
+            while root is not None:
+                root = root.next
+                n_nodes += 1
+            return n_nodes
+    
+        if headA is None or headB is None:
+            return None
+        n_A_nodes = get_n_nodes(headA)
+        n_B_nodes = get_n_nodes(headB)
+        n_diff_nodes = abs(n_A_nodes - n_B_nodes)
+        if n_A_nodes < n_B_nodes: # headA is longest
+            headA, headB = headB, headA
+        for _ in range(n_diff_nodes):
+            headA = headA.next
+        while headA and headB:
+            if headA is headB:
+                return headA
+            headA = headA.next
+            headB = headB.next
+        return None
+```