Linked List Cycle I & II

Given a linked list, determine if it has a cycle in it.

Follow up:
Can you solve it without using extra space?

使用快慢指针，快指针走两部，慢指针走一步，如果两者相同则有环。

public class Solution {
    //Time: O(n)  Space: O(1)
    public boolean hasCycle(ListNode head) {
        if (head == null || head.next == null) {
            return false;
        }
        
        ListNode fast = head;
        ListNode slow = head;
        while (fast.next != null && fast.next.next != null) {
            fast = fast.next.next;
            slow = slow.next;
            if (slow == fast) {
                return true;
            }
        }
        return false;
    }
}

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

Follow up:
Can you solve it without using extra space?
当判断有环后，让head和slow每次都一步，相遇即为环起点。

public class Solution {
    //Time: O(n)  Space: O(1)
    public ListNode detectCycle(ListNode head) {
        if (head == null || head.next == null) {
            return null;
        }
        ListNode fast = head;
        ListNode slow = head;
        while (fast.next != null && fast.next.next != null) {
            fast = fast.next.next;
            slow = slow.next;
            if (slow == fast) {
                break;
            }
        }
        if (fast.next == null || fast.next.next == null) {
            return null;
        }
        while (head != slow) {
            slow = slow.next;
            head = head.next;
        }
        return slow;
    }
}