Leetcode.com/problems/li…

Discuss:www.cnblogs.com/grandyang/p…

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).
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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.
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Example 3:

Input: head = [1], pos = -1
Output: false

Explanation: There is no cycle in the linked list.
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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?

Solution a:

Use HashSet to determine.

/** * 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? .next == null) { return false } val cacheSet = hashSetOf<ListNode>() var tempHead = head while (tempHead ! = null) { cacheSet.add(tempHead) if (cacheSet.contains(tempHead.next)) { return true } tempHead = tempHead.next } return  false } }Copy the code

Method 2:

No additional space can be used, i.e. the space complexity is O(1). This question is a classic interview question. The standard solution is to use two Pointers, one fast and one slow. If the fast pointer can catch up with the slow pointer, there is a ring, otherwise there is no ring.

class Solution { fun hasCycle(head: ListNode?) : Boolean { if (head? .next == null) { return false } var slowNode: ListNode? = head var quickNode: ListNode? = head while (slowNode ! = null && quickNode ! = null && quickNode.next ! = null) { slowNode = slowNode.next quickNode = quickNode.next? .next if (slowNode == quickNode) { return true } } return false } }Copy the code