# How to solve Leetcode 141. Linked List Cycle

## The two-pointers technique with linked list

## Problem statement

Given `head`

, the head of a linked list, determine if the linked list has a cycle in it.

Return `true`

if there is a cycle in the linked list. Otherwise, return `false`

.

### Example 1

```
Input: head = [3,2,0,-4], where -4 links next to 2.
Output: true
```

### Example 2

```
Input: head = [1,2], where 2 links next to 1.
Output: true
```

### Example 3

```
Input: head = [1], and 1 links to NULL.
Output: false
Explanation: There is no cycle in this linked list.
```

### Constraints

The number of the nodes in the list is in the range

`[0, 10^4]`

.`-10^5 <= Node.val <= 10^5`

.

**Follow up:** Can you solve it using `O(1)`

(i.e., constant) memory?

## Solution 1: Storing the visited nodes

### Code

```
#include <unordered_map>
#include <iostream>
struct ListNode {
int val;
ListNode *next;
ListNode(int x) : val(x), next(NULL) {}
};
bool hasCycle(ListNode *head) {
std::unordered_map<ListNode*, int> m;
while (head) {
if (m[head] > 0) {
return true;
}
m[head]++;
head = head->next;
}
return false;
}
int main() {
{
ListNode three(3);
ListNode two(2);
three.next = &two;
ListNode zero(0);
two.next = &zero;
ListNode four(4);
zero.next = &four;
four.next = &two;
std::cout << hasCycle(&three) << std::endl;
}
{
ListNode one(1);
ListNode two(2);
one.next = &two;
two.next = &one;
std::cout << hasCycle(&one) << std::endl;
}
{
ListNode one(1);
std::cout << hasCycle(&one) << std::endl;
}
}
```

```
Output:
1
1
0
```

### Complexity

Runtime:

`O(N)`

, where`N`

is the length of the linked list.Extra space:

`O(N)`

.

## Solution 2: Fast and slow runners

Imagine there are two runners both start to run along the linked list from the `head`

. One runs twice faster than the other.

If the linked list has a cycle in it, they will meet at some point. Otherwise, they never meet each other.

### Example 1

The slower runs `[3,2,0,-4,2,0,...]`

while the faster runs `[3,0,2,-4,0,2,...]`

. They meet each other at node `-4`

after three steps.

### Example 2

The slower runs `[1,2,1,2,...]`

while the faster runs `[1,1,1,...]`

. They meet each other at node `1`

after two steps.

### Code

```
#include <iostream>
struct ListNode {
int val;
ListNode *next;
ListNode(int x) : val(x), next(NULL) {}
};
bool hasCycle(ListNode *head) {
if (head == nullptr) {
return false;
}
ListNode* fast = head;
ListNode* slow = head;
while (fast && fast->next) {
fast = fast->next->next;
slow = slow->next;
if (fast == slow) {
return true;
}
}
return false;
}
int main() {
{
ListNode three(3);
ListNode two(2);
three.next = &two;
ListNode zero(0);
two.next = &zero;
ListNode four(4);
zero.next = &four;
four.next = &two;
std::cout << hasCycle(&three) << std::endl;
}
{
ListNode one(1);
ListNode two(2);
one.next = &two;
two.next = &one;
std::cout << hasCycle(&one) << std::endl;
}
{
ListNode one(1);
std::cout << hasCycle(&one) << std::endl;
}
}
```

```
Output:
1
1
0
```

### Complexity

Runtime:

`O(N)`

, where`N`

is the length of the linked list.Extra space:

`O(1)`

.

## References

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