The Floyd's Tortoise and Hare algorithm is a two-pointer technique that is used to detect cycles in a linked list. The approach involves using two pointers, one moving twice as fast as the other. If there's a cycle, the fast pointer will eventually meet the slow pointer. This method works in O(n) time with O(1) space.
Time Complexity: O(n), where n is the number of nodes in the list.
Space Complexity: O(1), constant space usage.
1#include <stdbool.h>
2
3struct ListNode {
4 int val;
5 struct ListNode *next;
6};
7
8bool hasCycle(struct ListNode *head) {
9 struct ListNode *slow = head, *fast = head;
10 while (fast != NULL && fast->next != NULL) {
11 slow = slow->next;
12 fast = fast->next->next;
13 if (slow == fast) {
14 return true;
15 }
16 }
17 return false;
18}In this solution, we create two pointers, slow and fast. The slow pointer moves one step at a time while the fast moves two steps. If the list contains a cycle, these pointers will eventually meet inside the cycle. If there's no cycle, fast or fast->next will become null.
This approach uses a HashSet to track the nodes visited during the traversal. If a node is encountered twice, there is a cycle. This method requires O(n) space but is simpler to understand.
Time Complexity: O(n^2) because of nested loops (non-optimal use of memory).
Space Complexity: O(n), where n is the number of nodes (uses additional storage).
1function ListNode(val) {
2 this.val = val;
3 this.next = null;
4}
5
6var hasCycle = function(head) {
7 const visited = new Set();
8 while (head) {
9 if (visited.has(head)) return true;
10 visited.add(head);
11 head = head.next;
12 }
13 return false;
14};JavaScript leverages a Set to remember nodes that have already been visited. Consequently, if a node reappears in the Set, a cycle is confirmed.