【LeetCode】2.Linked List — Linked List Cycle II 链表环2

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

To represent a cycle in the given linked list, we use an integer pos which represents the position (0-indexed) in the linked list where tail connects to. If pos is -1, then there is no cycle in the linked list.

Note: Do not modify the linked list.

 

Example 1:

Input: head = [3,2,0,-4], pos = 1
Output: tail connects to node index 1
Explanation: There is a cycle in the linked list, where tail connects to the second node.

Example 2:

Input: head = [1,2], pos = 0
Output: tail connects to node index 0
Explanation: There is a cycle in the linked list, where tail connects to the first node.

Example 3:

Input: head = [1], pos = -1
Output: no cycle
Explanation: There is no cycle in the linked list.

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

判断环的起始位置，我们还是设置两个不同步长的指针，步长相差为1，经理M步之后相遇，M即为圈的长度。然后我们将两个指针重置为head。其中一个p指针用于定位，另一个q用于遍历，当遍历指针经过M步遍历后回到p，则p为循环开始处。

/**
 * 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 *p=head;
        ListNode *q=head;
        int counter=0;
        while(p!=NULL&&p->next!=NULL)
        {
            p=p->next->next;
            q=q->next;
            counter++;
            if(p==q){
                p=head;
                q=head;
                while(1)
                {
                    for(int i=0;i<counter;i++) q=q->next;
                    if(p==q) return p;
                    else {
                        p=p->next;
                        q=p;
                    }
                }
            }
        }
         return NULL;
    }
};