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?

解題思路

設鏈表長度為n，頭結點與循環節點之間的長度為k。定義兩個指針slow和fast，slow每次走一步，fast每次走兩步。當兩個指針相遇時，有：

• fast = slow * 2
• fast - slow = (n - k)的倍數
  由上述兩個式子能夠得到slow為（n-k）的倍數

兩個指針相遇後，slow指針回到頭結點的位置，fast指針保持在相遇的節點。此時它們距離循環節點的距離都為k，然後以步長為1遍歷鏈表，再次相遇點即為循環節點的位置。

實現代碼

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */

 //Runtime:16 ms
class Solution {
public:
    ListNode *detectCycle(ListNode *head) {
        if
 
 (head == NULL)
        {
            return NULL;
        }

        ListNode *slow = head;
        ListNode *fast = head;
        while (fast->next && fast->next->next)
        {
            slow = slow->next;
            fast = fast->next->next;
            if (fast == slow)
            {
                break
 
;
            }
        }

        if (fast->next && fast->next->next)
        {
            slow = head;
            while (slow != fast)
            {
                slow = slow->next;
                fast = fast->next;
            }

            return slow;
        }

        return NULL;
    }
};

[LeetCode] Linked List Cycle II