[Leetcode] linked list cycle II

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

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

 

Idea: fast and slow pointer application. Time complexity O (N), space complexity O (1)

 1 /** 2  * Definition for singly-linked list. 3  * struct ListNode { 4  *     int val; 5  *     ListNode *next; 6  *     ListNode(int x) : val(x), next(NULL) {} 7  * }; 8  */ 9 class Solution {10 public:11     ListNode *detectCycle(ListNode *head) {12         if (head == NULL) return head;13         ListNode *fast = head;14         ListNode *slow = head;15         16         while (fast != NULL && fast->next != NULL) {17             fast = fast->next->next;18             slow = slow->next;19             if (fast == slow) break;20         }21         22         if (fast == NULL || fast->next == NULL) 23             return NULL;24             25         slow = head;26         while (slow != fast) {27             slow = slow->next;28             fast = fast->next;29         }30         return slow;31     }32 };

 

[Leetcode] linked list cycle II