(LeetCode) Linked List Cycle

Linked List Cycle

  • Explore : Interview > Top Interveiw Questions > Easy Collection
  • 분류 : Linked List
  • 난이도 : Easy

Problem

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

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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

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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

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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, 10^4].
  • -10^5 <= Node.val <= 10^5
  • pos is -1 or a valid index in the linked-list.

Solution 1 - Set

Exapnder
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/**
* Example:
* var li = ListNode(5)
* var v = li.`val`
* Definition for singly-linked list.
* class ListNode(var `val`: Int) {
* var next: ListNode? = null
* }
*/

class Solution {
fun hasCycle(head: ListNode?): Boolean {
if (head == null) {
return false
}
var set = HashSet<ListNode>()
var node = head
while (node?.next != null) {
if (set.contains(node)) {
return true
} else {
set.add(node)
node = node?.next
}
}
return false
}
}

Solution 2 - Next Pointer

Exapnder
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/**
* Example:
* var li = ListNode(5)
* var v = li.`val`
* Definition for singly-linked list.
* class ListNode(var `val`: Int) {
* var next: ListNode? = null
* }
*/

class Solution {
fun hasCycle(head: ListNode?): Boolean {
var front = head
var tail = head?.next?.next

while (tail != null && tail.next != null) {
front = front?.next
tail = tail?.next?.next
if (front == tail) {
return true
}
}
return false
}
}

Point of Thinking

  • Set에 모든 노드의 val을 넣고 검증하는 전통적인 방법.
  • next pointer를 이용해 비교해나가면서 순회하는 공간복잡도 최적의 방법.