142. Linked List Cycle II 环形链表 |

题目链接
tag:

• Medium；
• Two Pointers；

question：
  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.

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

思路：
  这个求单链表中的环的起始点是之前那个判断单链表中是否有环的延伸，可参之前那道Linked List Cycle |。这里还是要设快慢指针，不过这次要记录两个指针相遇的位置。然后让slow复制为头结点，然后fast和slow同时一步一步走，再相遇的位置就是链表中环的起始位置。

假设第一次相遇时slow走过的距离：a+b，fast走过的距离：a+b+c+b。

因为fast的速度是slow的两倍，所以fast走的距离是slow的两倍，有 2(a+b) = a+b+c+b，可以得到a=c（这个结论很重要！）。

代码如下：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *detectCycle(ListNode *head) {
        ListNode *slow = head, *fast = head;
        while (fast && fast->next) {
            fast = fast->next->next;
            slow = slow->next;
            if (slow == fast)
                break;
        }
        if (!fast || !fast->next)
            return NULL;
        slow = head;
        while (slow != fast) {
            slow = slow->next;
            fast = fast->next;
        }
        return slow;
    }
};