# 142 - Linked List Cycle II

解法一 - Fast & Slow Pointer

這個解法滿巧妙的，叫做 Floyd's Tortoise and Hare，詳細可以看 Leetcode 解答。

/**
 * 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;
        ListNode* fast = head;

        while(fast != nullptr && fast->next != nullptr) {
            fast = fast->next->next;
            slow = slow->next;

            if(slow == fast) {
                ListNode* tmp = head;
                while(tmp != slow) {
                    tmp = tmp->next;
                    slow = slow->next;
                }
                return tmp;
            }
        }

        return nullptr;
    }
};