Golden interview programmers - 02.08 loop detection face questions (speed pointer)

1. Topic

Given a list loop, the algorithm returns to realize a loop at the beginning of the node.
A cycloalkyl list definition: the next element in the list point to a node in the node preceding it had occurred, it indicates the presence of the chain loop.

示例 1：
输入：head = [3,2,0,-4], pos = 1
输出：tail connects to node index 1
解释：链表中有一个环，其尾部连接到第二个节点。

示例 2：
输入：head = [1,2], pos = 0
输出：tail connects to node index 0
解释：链表中有一个环，其尾部连接到第一个节点。

示例 3：
输入：head = [1], pos = -1
输出：no cycle
解释：链表中没有环。

进阶：
你是否可以不用额外空间解决此题？

Source: stay button (LeetCode)
link: https: //leetcode-cn.com/problems/linked-list-cycle-lcci
copyrighted by deduction from all networks. Commercial reprint please contact the authorized official, non-commercial reprint please indicate the source.

2. Problem Solving

• Pointer speed, fast take two steps, slow step, when fast==slowthe presence of a ring
• Then slow pointer back to square one, only one speed all go one step further, meet again point is the entrance to the ring
• DETAILED derivation, see: the list loop detection

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