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?
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 8 */ 9 class Solution { 10 public: 11 ListNode *detectCycle(ListNode *head) { 12 if(NULL == head || NULL == head->next) return NULL; 13 ListNode *p1 = head->next; 14 ListNode *p2 = head->next->next; 15 while(p2 != NULL && p2->next != NULL && p1 != p2){ 16 p1 = p1->next; 17 p2 = p2->next->next; 18 } 19 if(p1 == p2){ 20 ListNode *tp = head; 21 while(tp != p2){ 22 tp=tp->next; 23 p2=p2->next; 24 } 25 return tp; 26 } 27 else return NULL; 28 } 29 };
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
leetcode - Linked List Cycle II
原文：http://www.cnblogs.com/shnj/p/4646136.html