141. Linked List Cycle Leetcode Solution
Linked List Cycle Leetcode Problem :
Given head, the head of a linked list, determine if the linked list has a cycle in it.
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. Note that pos is not passed as a parameter.
Return true if there is a cycle in the linked list. Otherwise, return false.
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: true
Example 2:
- Input: head = [1], pos = -1
- Output: false
Intuition :
Use a Floyd’s Tortoise and Hare algorithm to detect a cycle in a linked list and find the node where the cycle starts.
Approach :
Use Floyd’s Tortoise and Hare algorithm. it is a widely used algorithm for detecting cycles in a linked list
1 – Initialize two pointers, slow and fast, both pointing to the head of the linked list.
2 – Move the slow one step at a time, and the fast two steps at a time. This simulates a slow and fast movement through the list.
3 – If there is no cycle in the list, the ha fast will reach the end of the list, and you can stop the algorithm.
4 – If there is a cycle in the list, the fast will eventually catch up to the slow because it’s moving twice as fast. When this happens, you have detected a cycle.
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Code :
class Solution {
public:
bool hasCycle(ListNode *head) {
if(head==NULL) return false; // for empty linked list
ListNode * slow=head;
ListNode * fast=head;
// By checking both conditions, you ensure that the fast pointer can move ahead by two nodes (fast movement), and the slow pointer can move ahead by one node (slow movement) without encountering null pointers.
while(fast->next!=nullptr && fast->next->next!=nullptr){
slow=slow->next;
fast=fast->next->next;
if(slow==fast){
return true;
}
}
return false;
}
};
public class Solution {
public boolean hasCycle(ListNode head) {
ListNode tortoise = head;
ListNode rabbit = head;
while (rabbit != null && rabbit.next != null) {
// Tortoise moves one step
tortoise = tortoise.next;
// Rabbit moves two steps
rabbit = rabbit.next.next;
// Check if they meet at the same spot
if (tortoise == rabbit) {
return true;
}
}
return false;
}
}
class Solution:
def hasCycle(self, head: Optional[ListNode]) -> bool:
address = []
curr = head
while(curr is not None):
address.append(curr)
curr = curr.next
if(curr in address):
return True
return False
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