Given a linked list, return the node where the cycle begins. If there is no cycle, return null.
Follow up:
Can you solve it without using extra space?
前一篇介绍了如何寻找到环,在找到环后可以立马将移动比较快的指针恢复到头指针处,并以后每次向前移动一步。当它们的值相同时即表示是环的入口。
runtime:13ms
/**
* 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) {
if(head==NULL) return NULL;
ListNode * slow=head;
ListNode * fast=head;
while(slow->next!=NULL&&fast->next!=NULL&&fast->next->next!=NULL)
{
slow=slow->next;
fast=fast->next->next;
//如果找到了环,将fast指针重新设置为head,并每次移动一步,当它再次和slow相同时就是环的起点
if(slow==fast)
{
fast=head;
while(slow!=fast)
{
slow=slow->next;
fast=fast->next;
}
return fast;
}
}
return NULL;
}
};