# 142. Linked List Cycle II

Given a linked list, return the node where the cycle begins. If there is no cycle, return `null`.

To represent a cycle in the given linked list, we use an integer `pos` which represents the position (0-indexed) in the linked list where tail connects to. If `pos` is `-1`, then there is no cycle in the linked list.

Note: 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.
``````

Can you solve it without using extra space?

``````class Solution {
public:
ListNode *detectCycle(ListNode *head) {
ListNode *slow = head, *fast = head;
while (fast && fast->next) {
slow = slow->next;
fast = fast->next->next;
if (slow == fast) break;
}
if (!fast || !fast->next) return NULL;
while (slow != fast) {
slow = slow->next;
fast = fast->next;
}
return fast;
}
};
``````

Github 同步地址：

https://github.com/grandyang/leetcode/issues/142

Find the Duplicate Number