Linked List Cycle II

描述


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



Note: Do not modify the linked list.


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

题解

/**
 * 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 *fast = head, *slow = head, *entry = head;
        while(fast != nullptr && fast->next != nullptr) {
            slow = slow->next;
            fast = fast->next->next;
            if (fast == slow) {
                while(slow != entry) {
                    slow  = slow->next;
                    entry = entry->next;
                }
                return entry;
            }
        }
        return NULL;
    }
};