Intersection of Two Linked Lists

Author: gavinfish

160 Intersection of Two Linked Lists.py

@@ -0,0 +1,72 @@
+'''
+Write a program to find the node at which the intersection of two singly linked lists begins.
+
+
+For example, the following two linked lists:
+
+A:          a1 → a2
+                   ↘
+                     c1 → c2 → c3
+                   ↗            
+B:     b1 → b2 → b3
+begin to intersect at node c1.
+
+
+Notes:
+
+If the two linked lists have no intersection at all, return null.
+The linked lists must retain their original structure after the function returns.
+You may assume there are no cycles anywhere in the entire linked structure.
+Your code should preferably run in O(n) time and use only O(1) memory.
+'''
+
+# Definition for singly-linked list.
+class ListNode(object):
+    def __init__(self, x):
+        self.val = x
+        self.next = None
+
+
+class Solution(object):
+    def getIntersectionNode(self, headA, headB):
+        """
+        :type head1, head1: ListNode
+        :rtype: ListNode
+        """
+        nodeA, nodeB = headA, headB
+        while nodeA != nodeB:
+            nodeA = nodeA.next if nodeA else headB
+            nodeB = nodeB.next if nodeB else headA
+        return nodeA
+
+    def getIntersectionNode_diff(self, headA, headB):
+        """
+        :type head1, head1: ListNode
+        :rtype: ListNode
+        """
+
+        def get_length(node):
+            length = 0
+            while node:
+                node = node.next
+                length += 1
+            return length
+
+        len1 = get_length(headA)
+        len2 = get_length(headB)
+        if len1 > len2:
+            for __ in range(len1 - len2):
+                headA = headA.next
+        else:
+            for __ in range(len2 - len1):
+                headB = headB.next
+        while headA:
+            if headA == headB:
+                return headA
+            headA = headA.next
+            headB = headB.next
+        return None
+
+
+if __name__ == "__main__":
+    None