leetcode - Linked List Cycle II

leetcode - Linked List Cycle II

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?

/**
* 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(NULL == head || NULL == head->next) return NULL;
ListNode *p1 = head->next;
ListNode *p2 = head->next->next;
while(p2 != NULL && p2->next != NULL && p1 != p2){
p1 = p1->next;
p2 = p2->next->next;
}
if(p1 == p2){
ListNode *tp = head;
while(tp != p2){
tp=tp->next;
p2=p2->next;
}
return tp;
}
else return NULL;
}
};

两个指针ptr1和ptr2，都从链表头开始走，ptr1每次走一步，ptr2每次走两步，等两个指针重合时，就说明有环，否则没有。如果有环的话，那么让ptr1指向链表头，ptr2不动，两个指针每次都走一步，当它们重合时，所指向的节点就是环开始的节点。
http://blog.sina.com.cn/s/blog_6f611c300101fs1l.html#cmt_3246993