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.

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

思路

先判断是否有环

有环：slow节点从头开始遍历，fast节点从而且相遇的地方开始遍历，等到相遇的节点即为相遇点。

无还：返回NULL

c++11

/**
 * 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) {
            slow = slow->next;
            fast = fast->next->next;
            if (slow == fast) {
                slow = head;
                while (slow != fast) {
                    slow = slow->next;
                    fast = fast->next;
                }
                break;
            }
        }
        if (!fast || !fast->next) return NULL;
        else return fast;
    }
};

ac结果：

Runtime: 20 ms, faster than 30.50% of C++ online submissions for Linked List Cycle II.
Memory Usage: 9.7 MB, less than 63.00% of C++ online submissions for Linked List Cycle II.