Suppose we have two sorted linked lists L1 and L2, we have to make a new sorted linked list which contains the intersection of these two lists.
So, if the input is like L1 = [2, 4, 8] L2 = [3, 4, 8, 10], then the output will be [4, 8, ]
To solve this, we will follow these steps −
- head := a new node with value 0
- cur := head
- while l1 and l2 are not empty, do
- if value of l1 < value of l2, then
- l1 := next of l1
- otherwise when value of l2 < value of l1, then
- l2 := next of l2
- otherwise,
- next of cur := a new node with value same as value of l1
- l1 := next of l1
- l2 := next of l2
- cur := next of cur
- if value of l1 < value of l2, then
- return next of head
Let us see the following implementation to get better understanding −
Example
class ListNode:
def __init__(self, data, next = None):
self.val = data
self.next = next
def make_list(elements):
head = ListNode(elements[0])
for element in elements[1:]:
ptr = head
while ptr.next:
ptr = ptr.next
ptr.next = ListNode(element)
return head
def print_list(head):
ptr = head
print('[', end = "")
while ptr:
print(ptr.val, end = ", ")
ptr = ptr.next
print(']')
class Solution:
def solve(self, l1, l2):
head = cur = ListNode(0)
while l1 and l2:
if l1.val < l2.val:
l1 = l1.next
elif l2.val < l1.val:
l2 = l2.next
else:
cur.next = ListNode(l1.val)
l1 = l1.next
l2 = l2.next
cur = cur.next
return head.next
ob = Solution()
L1 = make_list([2, 4, 8])
L2 = make_list([3, 4, 8, 10])
print_list(ob.solve(L1, L2))Input
[2, 4, 8], [3, 4, 8, 10]
Output
[4, 8, ]