Algorithm
Problem Name: 141. Linked List Cycle
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:
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:
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:
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 <= 105posis-1or a valid index in the linked-list.
Code Examples
#1 Code Example with C Programming
Code -
C Programming
bool hasCycle(struct ListNode *head) {
struct ListNode *n, *nn;
if (!head) return false;
n = nn = head;
do {
n = n->next; // one step
nn = nn->next;
if (nn) nn = nn->next; // two steps
} while (nn && n != nn);
if (nn) return true;
return false;
}
Input
Output
#2 Code Example with C++ Programming
Code -
C++ Programming
class Solution {
public:
bool hasCycle(ListNode *head) {
auto one = head, two = head;
while(two && two->next){
one = one->next;
two = two->next->next;
if(one == two) return true;
}
return false;
}
};
Input
Output
#3 Code Example with Java Programming
Code -
Java Programming
public class Solution {
public boolean hasCycle(ListNode head) {
ListNode slow = head;
ListNode fast = head;
while (fast != null && fast.next != null) {
slow = slow.next;
fast = fast.next.next;
if (slow == fast) {
return true;
}
}
return false;
}
}
Input
Output
#4 Code Example with Javascript Programming
Code -
Javascript Programming
const hasCycle = function(head) {
const seen = []
while(head != null) {
if(seen.indexOf(head) !== -1) {
return true
} else {
seen.push(head)
}
head = head.next
}
return false
};
Input
Output
#5 Code Example with Python Programming
Code -
Python Programming
class Solution:
def hasCycle(self, head: ListNode) -> bool:
slow, fast = head, head.next if head else None
while slow != None and fast != None:
if slow == fast:
return True
slow, fast = slow.next, fast.next.next if fast.next else None
return False
Input
Output
#6 Code Example with C# Programming
Code -
C# Programming
namespace LeetCode
{
public class _0141_LinkedListCycle
{
public bool HasCycle(ListNode head)
{
if (head == null || head.next == null) return false;
ListNode slow = head, fast = head.next;
while (slow != fast)
{
if (fast == null || fast.next == null) return false;
slow = slow.next;
fast = fast.next.next;
}
return true;
}
}
}
Input
Output