# Leetcode: Insert into a Cyclic Sorted List

Insert into a Cyclic Sorted List

Similar Problems:

Given a node from a cyclic linked list which is sorted in ascending order, write a function to insert a value into the list such that it remains a cyclic sorted list. The given node can be a reference to any single node in the list, and may not be necessarily the smallest value in the cyclic list.

If there are multiple suitable places for insertion, you may choose any place to insert the new value. After the insertion, the cyclic list should remain sorted.

If the list is empty (i.e., given node is null), you should create a new single cyclic list and return the reference to that single node. Otherwise, you should return the original given node.

In the figure above, there is a cyclic sorted list of three elements. You are given a reference to the node with value 3, and we need to insert 2 into the list.

The new node should insert between node 1 and node 3. After the insertion, the list should look like this, and we should still return node 3.

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

• Solution:
```## Blog link: https://code.dennyzhang.com/insert-into-a-cyclic-sorted-list
## Basic Ideas:
##    Find the smallest value which is no less than the target
##    If there are multiple suitable places for insertion,
##      here we choose to insert into the first suitable place after head.
##
## Question: how we know we are running the loop again?
##
## Complexity: Time O(n), Space O(1)
"""
# Definition for a Node.
class Node:
def __init__(self, val, next):
self.val = val
self.next = next
"""
class Solution:
"""
:type insertVal: int
:rtype: Node
"""
node = Node(insertVal, None)
# empty
node.next = node
return node

# one node

# find the smallest value, which is no less than the target
while True:
# end of the loop
if p.val > p.next.val:
# biggest or smallest
if insertVal >= p.val or insertVal <= p.next.val:
break

# should keep going
if insertVal > p.next.val and insertVal < p.val:
p = p.next
continue
break

if insertVal >= p.val and insertVal <= p.next.val:
break
p = p.next