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.
1function ListNode(val) {
2 this.val = val;
3 this.next = null;
4}
5
6var hasCycle = function(head) {
7 let slow = head, fast = head;
8 while (fast && fast.next) {
9 slow = slow.next;
10 fast = fast.next.next;
11 if (slow === fast) return true;
12 }
13 return false;
14};The JavaScript function leverages the same logic of two pointers. Different speeds of traversal will determine if a cycle exists when the pointers align.
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).
1#include <stdbool.h>
2#include <stdlib.h>
3
4struct ListNode {
5 int val;
6 struct ListNode *next;
7};
8
9bool hasCycle(struct ListNode *head) {
10 struct ListNode **nodes = malloc(10000 * sizeof(struct ListNode*));
11 int idx = 0;
12 while (head != NULL) {
13 for (int i = 0; i < idx; i++) {
14 if (nodes[i] == head) {
15 free(nodes);
16 return true;
17 }
18 }
19 nodes[idx++] = head;
20 head = head->next;
21 }
22 free(nodes);
23 return false;
24}This C solution simulates a HashSet using an array to record visited nodes. By checking existing entries, it detects cycles but uses O(n) space.