Insert into a Cyclic Sorted List

Given a node from a cyclic linked list which has been sorted, write a function to insert a value into the list such that it remains a cyclic sorted list. The given node can be any single node in the list. Return the inserted new node.

 Notice

3->5->1 is a cyclic list, so 3 is next node of 1.
3->5->1 is same with 5->1->3

Have you met this question in a real interview? 

Yes

Example

Given a list, and insert a value 4:
3->5->1

Return 5->1->3->4

java

/** * Definition for ListNode * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { *         val = x; *         next = null; *     } * } */public class Solution {    /*     * @param node: a list node in the list     * @param x: An integer     * @return: the inserted new list node     */    public ListNode insert(ListNode node, int x) {        // write your code here        ListNode n = new ListNode(x);        if (node == null) {            n.next = n;            return n;        }        ListNode curr = node.next;        ListNode prev = node;        while (curr != node) {            if (prev.val <= x && curr.val >= x) {                break;            }            if (prev.val > curr.val && (prev.val < x || curr.val > x)) {                break;            }            curr = curr.next;            prev = prev.next;        }        prev.next = n;        n.next = curr;        return node;    }}

python

"""Definition of ListNodeclass ListNode(object):    def __init__(self, val, next=None):        self.val = val        self.next = next"""class Solution:    """    @param: node: a list node in the list    @param: x: An integer    @return: the inserted new list node    """    def insert(self, node, x):        # write your code here        n = ListNode(x)        if node is None:            n.next = n            return n        prev, curr = node, node.next        while curr != node:            if prev.val <= x and curr.val >= x:                break            if prev.val > curr.val and (prev.val < x or curr.val > x):                break            curr, prev = curr.next, prev.next        prev.next = n        n.next = curr        return node