142 Linked List Cycle II Leetcode Solution
Linked List Cycle II Leetcode Problem :
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.
Linked List Cycle || Leetcode Solution :
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.
Example 1:
- Input: head = [1,2], pos = 0
- Output: tail connects to node index 0
Example 2:
- Input: head = [1], pos = -1
- Output: no cycle
Intuition :
Use a Floyd’s Cycle-Finding algorithm to detect a cycle in a linked list and find the node where the cycle starts.
Approach :
- When the two pointers meet, we know that there is a cycle in the linked list.
- We then reset the slow pointer to the head of the linked list and move both pointers at the same pace, one step at a time, until they meet again.
- The node where they meet is the starting point of the cycle.
- If there is no cycle in the linked list, the algorithm will return null.
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Code :
class Solution {
public:
ListNode* detectCycle(ListNode* head) {
ListNode* slow = head;
ListNode* fast = head;
while (fast && fast->next) {
slow = slow->next;
fast = fast->next->next;
if (slow == fast) {
slow = head;
while (slow != fast) {
slow = slow->next;
fast = fast->next;
}
return slow;
}
}
return nullptr;
}
};
public class Solution {
public ListNode detectCycle(ListNode head) {
ListNode slow = head;
ListNode fast = head;
while (fast != null && fast.next != null) {
slow = slow.next;
fast = fast.next.next;
if (slow == fast) {
slow = head;
while (slow != fast) {
slow = slow.next;
fast = fast.next;
}
return slow;
}
}
return null;
}
}
class Solution:
def detectCycle(self, head: ListNode) -> ListNode:
slow = head
fast = head
while fast and fast.next:
slow = slow.next
fast = fast.next.next
if slow == fast:
slow = head
while slow != fast:
slow = slow.next
fast = fast.next
return slow
return None
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