# 19. \[Linked List]: Linked List Cycle

**141. Linked List Cycle**

**Easy**

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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`.

**Example 1:**

!<https://assets.leetcode.com/uploads/2018/12/07/circularlinkedlist.png>

```
Input: head = [3,2,0,-4], pos = 1
Output: true
Explanation: There is a cycle in the linked list, where the tail connects to the 1st node (0-indexed).

```

**Example 2:**

!<https://assets.leetcode.com/uploads/2018/12/07/circularlinkedlist_test2.png>

```
Input: head = [1,2], pos = 0
Output: true
Explanation: There is a cycle in the linked list, where the tail connects to the 0th node.

```

**Example 3:**

!<https://assets.leetcode.com/uploads/2018/12/07/circularlinkedlist_test3.png>

```
Input: head = [1], pos = -1
Output: false
Explanation: There is no cycle in the linked list.

```

**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.

**Follow up:** Can you solve it using `O(1)` (i.e. constant) memory?

{% embed url="<https://leetcode.com/problems/linked-list-cycle/>" %}

### `With Array`

> time: O(n^2)

> space: O(n);

```jsx
var hasCycle = function(head) {
    
    const arr = [];
    let curr = head;

    while (curr !== null) {
        if (arr.includes(curr)) 
					return true;
        arr.push(curr);
        curr = curr.next;
    }

    return false;
};
```

### `With Set`

> time: O(n)

> time: O(n)

```jsx
var hasCycle = function(head) {
    
    const set = new Set();
    let curr = head;

    while (curr !== null) {
        if (set.has(curr)) 
					return true;
        set.add(curr);
        curr = curr.next;
    }

    return false;
};
```

### `With two pointers`

> time: O(n)

> time: O(1)

```jsx
var hasCycle = function(head) {
    if (head === null) return false;
    if (head.next === null) return false;
    
    let slow = head;
    let fast = head.next.next;

    while (fast !== null && fast.next !== null) {
        if (fast === slow) return true;
        slow = slow.next;
        fast = fast.next.next;
    }

    return false;
};
```