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?
分析:
假设有环,链表分两部分,环外部分和环内部分,环外部分长为L(是边的长度,不是点的个数),环内长为S,设置快慢指针,相遇点在环内距离起始点T处,那么快指针和慢指针之间有如下关系:L+m*S+T = 2*(L+n*S+T), 化简得(m-2*n-1)*S+(S-T) = L, 那么一个指针从链表起始点出发,另一个指针从相遇点出发,二者会在环的起始点相遇。
/**
* 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||head->next == NULL)
return NULL;
ListNode *fast = head, *slow = head;
while(fast != NULL)
{
fast = fast->next;
if(fast!=NULL)fast = fast->next;
else
return NULL;
slow = slow->next;
if(fast == slow)
break;
}
if(fast == NULL)
return NULL;
ListNode *p = head;
ListNode *q = fast;
while(p!=q)
{
p = p->next;
q = q->next;
}
return p;
}
};