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# Linked List Cycle

ID: 141; easy

/**

* Definition for singly-linked list.

* type ListNode struct {

* Val int

* Next *ListNode

* }

*/

func hasCycle(head *ListNode) bool {

m := make(map[*ListNode]bool)

for head != nil {

if _,found := m[head]; found {

return true

}

m[head] = true

head = head.Next

}

return false

}

/**

* Definition for singly-linked list.

* type ListNode struct {

* Val int

* Next *ListNode

* }

*/

func hasCycle(head *ListNode) bool {

tortoise, hare := head, head

for tortoise != nil && hare != nil && hare.Next != nil {

tortoise = tortoise.Next

hare = hare.Next.Next

if tortoise == hare {

return true

}

}

return false

}

/**

* Definition for ListNode

* public class ListNode {

* int val;

* ListNode next;

* ListNode(int x) {

* val = x;

* next = null;

* }

* }

*/

public class Solution {

/**

* @param head: The first node of linked list.

* @return: True if it has a cycle, or false

*/

public boolean hasCycle(ListNode head) {

if (head == null || head.next == null)

return false;

ListNode slow = head;

ListNode fast = head;

while (fast.next != null && fast.next.next != null) {

slow = slow.next;

fast = fast.next.next;

if (slow == fast) return true;

}

return false;

}

}

The tortoise move by 1 step each time and the hare moves by 2 steps at a time. If there is a cycle, the hare will eventually catch the tortoise at a position.

A simple idea of the proof can be tracking the

**gap**between the tortoise and the hare. By construction, the gap increases by 1 each iteration. Eventually, the gap with become n, where n is the number of elements in the**cycle**(not the whole list). This is the time when the tortoise and the hare meet. More proofs:Last modified 2yr ago