#160: Intersection of Two Linked Lists

Author: papilo-cloud

Linkedlist/intersectn-of-two-lnkdlst.py

@@ -0,0 +1,90 @@
+# Given the heads of two singly linked-lists headA and headB, return the node
+# at which the two lists intersect. If the two linked lists have no intersection
+# at all, return null.
+
+# For example, the following two linked lists begin to intersect at node c1:
+
+
+# The test cases are generated such that there are no cycles anywhere in the 
+# entire linked structure.
+
+# Note that the linked lists must retain their original structure after the 
+# function returns.
+
+# Custom Judge:
+# The inputs to the judge are given as follows (your program is not given these inputs):
+
+# intersectVal - The value of the node where the intersection occurs. This is 0 if 
+# there is no intersected node.
+
+# listA - The first linked list.
+# listB - The second linked list.
+# skipA - The number of nodes to skip ahead in listA (starting from the head) to get
+# to the intersected node.
+# skipB - The number of nodes to skip ahead in listB (starting from the head) to get
+# to the intersected node.
+# The judge will then create the linked structure based on these inputs and pass the
+# two heads, headA and headB to your program. If you correctly return the intersected
+# node, then your solution will be accepted.
+
+
+# Example 1:
+# Input: intersectVal = 8, listA = [4,1,8,4,5], listB = [5,6,1,8,4,5], skipA = 2, skipB = 3
+# Output: Intersected at '8'
+# Explanation: The intersected node's value is 8 (note that this must not be 0 if the two
+# lists intersect).
+# From the head of A, it reads as [4,1,8,4,5]. From the head of B, it reads as [5,6,1,8,4,5].
+# There are 2 nodes before the intersected node in A; There are 3 nodes before the intersected node in B.
+# - Note that the intersected node's value is not 1 because the nodes with value 1 in A and B
+# (2nd node in A and 3rd node in B) are different node references. In other words, they point
+# to two different locations in memory, while the nodes with value 8 in A and B 
+# (3rd node in A and 4th node in B) point to the same location in memory.
+
+# Example 2:
+# Input: intersectVal = 2, listA = [1,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1
+# Output: Intersected at '2'
+# Explanation: The intersected node's value is 2 (note that this must not be 0 if the 
+# two lists intersect).
+# From the head of A, it reads as [1,9,1,2,4]. From the head of B, it reads as [3,2,4].
+# There are 3 nodes before the intersected node in A; There are 1 node before the intersected
+# node in B.
+
+# Example 3:
+# Input: intersectVal = 0, listA = [2,6,4], listB = [1,5], skipA = 3, skipB = 2
+# Output: No intersection
+# Explanation: From the head of A, it reads as [2,6,4]. From the head of B, it reads as [1,5].
+# Since the two lists do not intersect, intersectVal must be 0, while skipA and skipB can be 
+# arbitrary values.
+# Explanation: The two lists do not intersect, so return null.
+ 
+
+# Constraints:
+
+# The number of nodes of listA is in the m.
+# The number of nodes of listB is in the n.
+# 1 <= m, n <= 3 * 104
+# 1 <= Node.val <= 105
+# 0 <= skipA <= m
+# 0 <= skipB <= n
+# intersectVal is 0 if listA and listB do not intersect.
+# intersectVal == listA[skipA] == listB[skipB] if listA and listB intersect.
+
+
+
+from typing import Optional
+
+# Definition for singly-linked list.
+class ListNode:
+    def __init__(self, x):
+        self.val = x
+        self.next = None
+
+class Solution:
+    def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> Optional[ListNode]:
+        a = headA
+        b = headB
+
+        while a != b:
+            a = a.next if a else headB
+            b = b.next if b else headA
+        return a