[数据结构与算法]708. Insert into a Sorted Circular Linked List. Sol

Given a Circular Linked List node, which is sorted in ascending order, write a function to insert a value?insertVal?into the list such that it remains a sorted circular list. The given node can be a reference to any single node in the list and may not necessarily be the smallest value in the circular list.

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

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

Example 1:

?

Input: head = [3,4,1], insertVal = 2
Output: [3,4,1,2]
Explanation: In the figure above, there is a sorted circular 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 be inserted between node 1 and node 3. After the insertion, the list should look like this, and we should still return node 3.

Example 2:

Input: head = [], insertVal = 1
Output: [1]
Explanation: The list is empty (given head is?null). We create a new single circular list and return the reference to that single node.

Example 3:

Input: head = [1], insertVal = 0
Output: [1,0]

Constraints:

• The number of nodes in the list is in the range?[0, 5 * 104].
• -106?<= Node.val, insertVal <= 106

  class Solution:
      def insert(self, head, insertVal: int):
          insert = ListNode()
          insert.val = insertVal
          if head == None:
              insert.next = insert
              return insert
          elif head.next == head:
              head.next = insert
              insert.next = head
              return head
          else:
              ptr = head
              while ptr.next.val-ptr.val >= 0:
                  ptr = ptr.next
                  if ptr ==  head:
                      break
              large = ptr
              small = ptr.next
              if insertVal <= small.val or insertVal>= large.val:
                  large.next = insert
                  insert.next = small
              else:
                  while small.next.val < insertVal:
                      small = small.next
                  temp = small.next
                  small.next = insert
                  insert.next = temp
              return head