/**
Given the head of a linked list, return the node where the cycle begins. If
there is no cycle, return null.
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 (0-
indexed). It is -1 if there is no cycle. Note that pos is not passed as a
parameter.
Do not modify the linked list.
Example 1:
Input: head = [3,2,0,-4], pos = 1
Output: tail connects to node index 1
Explanation: There is a cycle in the linked list, where tail connects to the
second node.
Example 2:
Input: head = [1,2], pos = 0
Output: tail connects to node index 0
Explanation: There is a cycle in the linked list, where tail connects to the
first node.
Example 3:
Input: head = [1], pos = -1
Output: no cycle
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?
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*/
//leetcode submit region begin(Prohibit modification and deletion)
/**
* Definition for singly-linked list.
* class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
public class Solution {
public ListNode detectCycle(ListNode head) {
ListNode slow = head, fast = head;
int fastLen = 0, slowLen = 0;
Set<ListNode> nodeSet = new HashSet<>();
while (head != null && fast != null) {
fast = fast.next;
fastLen++;
if (fast == slow) {
break;
}
if (fast != null) {
fast = fast.next;
fastLen++;
}
if (fast == slow) {
break;
}
slow = slow.next;
slowLen++;
}
if (fast == null || slow == null) {
return null;
}
for (int i = 0; i < fastLen - slowLen; i++) {
nodeSet.add(fast);
fast = fast.next;
}
slow = head;
while (!nodeSet.contains(slow)) {
slow = slow.next;
}
return slow;
}
}
//leetcode submit region end(Prohibit modification and deletion)
