19. [Linked List]: Linked List Cycle
141. Linked List Cycle
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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
is1
or a valid index in the linked-list.
Follow up: Can you solve it using O(1)
(i.e. constant) memory?
With Array
With Array
time: O(n^2)
space: O(n);
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
With Set
time: O(n)
time: O(n)
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
With two pointers
time: O(n)
time: O(1)
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;
};
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