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).
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 <= 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?
解法一:
使用 HashSet 来判断。
/**
* 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
}
}
解法二:
不能使用额外的空间,即空间复杂度是 O(1)。该问题是经典面试问题,其标准解法是用两个指针,一快一慢,如果在快的指针能够追上慢的指针,则有环,否则无环。
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
}
}