Table of Contents
Starting node of cycle in Linked List
Problem Statement
Given the head
of a linked list, return the node where the cycle begins. If there is no cycle, return null
.
There is a cycle in a linked list if there is some node in the list that can be reached again by continuously following the next
pointer. Internally, pos
is used to denote the index of the node that tail’s next
pointer is connected to (0-indexed). It is -1
if there is no cycle. Note that pos
is not passed as a parameter.
Do not modify the linked list.
Example 1:
Input: head = [3,2,0,-4], pos = 1 Output: tail connects to node index 1 Explanation: There is a cycle in the linked list, where tail connects to the second node.
Example 2:
Input: head = [1,2], pos = 0 Output: tail connects to node index 0 Explanation: There is a cycle in the linked list, where tail connects to the first node.
Example 3:
Input: head = [1], pos = -1 Output: no cycle Explanation: There is no cycle in the linked list.
Constraints:
- The number of the nodes in the list is in the range
[0, 104]
. -105 <= Node.val <= 105
pos
is-1
or a valid index in the linked-list.
Follow up: Can you solve it using O(1)
(i.e. constant) memory?
Hashing Technique
Complexity
- Time Complexity: O(n).
- Space Complexity: O(n).
/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func detectCycle(head *ListNode) *ListNode {
if head == nil || head.Next == nil {
return nil
}
temp := head
// define map
seen := map[*ListNode]bool{}
for temp != nil {
// we found the cycle and this is the first node in cycle
if seen[temp] == true {
return temp
}
seen[temp] = true
// move to next position
temp = temp.Next
}
return nil
}
Floyd’s Cycle Detection Algorithm
Fast and Slow Pointer Technique
Complexity
- Time Complexity: O(n).
- Space Complexity: O(1).
/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func detectCycle(head *ListNode) *ListNode {
if head == nil || head.Next == nil {
return nil
}
slow, fast := head, head
for fast != nil && fast.Next != nil {
slow = slow.Next
// This is critical condition to move fast by 2 positions
fast = fast.Next.Next
// we have detected a cycle
if slow == fast {
// now find the first node of a cycle
slow = head
for slow != fast {
fast = fast.Next
slow = slow.Next
}
return slow
}
}
return nil
}
Please visit https: https://codeandalgo.com for more such contents.
Leave a Reply