【Lintcode】103.Linked List Cycle II

题目：

Given a linked list, return the node where the cycle begins.

If there is no cycle, return null.

Example

Given -21->10->4->5, tail connects to node index 1，return 10

 

题解：

Solution 1 ()

class Solution {
public:
    ListNode *detectCycle(ListNode *head) {
        if (!head) {
            return head;
        }
        
        ListNode* slow = head;
        ListNode* fast = head;
        while (fast && fast->next) {
            slow = slow->next;
            fast = fast->next->next;
            if (slow == fast) {
                break;
            }
        }
        if (!fast || !fast->next) {
            return nullptr;
        }
        
        slow = head;
        while (slow != fast) {
            slow = slow->next;
            fast = fast->next;
        }
        
        return slow;
    }
};