LeetCode 708: Insert into a Sorted Circular Linked List Solution
Master LeetCode problem 708 (Insert into a Sorted Circular Linked List), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.
708. Insert into a Sorted Circular Linked List
Problem Explanation
Explanation
To insert a node into a sorted circular linked list, we need to find the correct position where the new node should be inserted while maintaining the sorted order. We can start by checking if the list is empty. If the list is empty, we can simply create a new node and set it as the head. If the list is not empty, we iterate through the list to find the correct position where the new node should be inserted.
Our approach will be to iterate through the list and check if the value of the current node is less than the value of the next node. If it is, then the new node should be inserted between these two nodes. If the value of the current node is greater than the value of the next node, it means we have reached the end of the circular list and the new node should be inserted at this point.
After finding the correct position, we update the pointers to insert the new node. We need to handle the edge case where the new node should be inserted at the beginning of the list. In this case, we update the head pointer accordingly.
The time complexity of this approach is O(n) where n is the number of nodes in the linked list, as we may need to iterate through the entire list to find the correct position to insert the new node. The space complexity is O(1) as we are not using any extra data structures.
Solution Code
class Solution {
public Node insert(Node head, int insertVal) {
if (head == null) {
Node newNode = new Node(insertVal, null);
newNode.next = newNode;
return newNode;
}
Node prev = head;
Node curr = head.next;
while (curr != head) {
if (prev.val <= insertVal && insertVal <= curr.val) {
break;
}
if (prev.val > curr.val && (insertVal >= prev.val || insertVal <= curr.val)) {
break;
}
prev = curr;
curr = curr.next;
}
prev.next = new Node(insertVal, curr);
return head;
}
}Try It Yourself
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