LeetCode 142. Linked List Cycle II

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.

第一步：

1. 设置一个慢指针，一次走一步

2. 设置一个快指针，一次走两步

3. 如果慢指针和快指针都指向同一个地址，则表明为这是一个环。

4. 否则，无环

第二步：

1. L1：头结点和入口结点之间的距离

2. L2：入口结点和相遇结点之间的距离

3. C：环的长度

4. n：快指针绕环一周的次数（当第一次，快慢指针相遇）

===> 1. 当相遇，慢指针经过的总距离：L1+L2

===> 2. 当相遇，快指针经过的总距离：L1+L2 + n*C

===> 头部位置和入口位置之间的距离=相遇位置（fast）和入口位置（entry）向前移动之间的距离。

class Solution 
{ 
public:
    ListNode *detectCycle(ListNode *head)
    {
        if (head == NULL || head->next == NULL) return NULL;
        
        ListNode *slow = head;  
        ListNode *fast = head;  
        bool isCycle = false;   

        while (slow != NULL && fast != NULL)
        {
            slow = slow->next;      
            if (fast->next == NULL) return NULL;
            fast = fast->next->next;
            if (slow == fast)    
            {
                isCycle = true;    
                break;
            }
        }

        if (!isCycle) return NULL; 
        slow = head;
        while (slow != fast)    
        {
            slow = slow->next;      
            fast = fast->next;      
        }
    
        return slow;

    }
};