标签：bsp head cycle col can lower nod node return
Given a linked list, return the node where the cycle begins. If there is no cycle, return null.
Note: Do not modify the linked list.
Follow up:
Can you solve it without using extra space?
头结点到cycle begins的点 距离是A, cycle begins的点 到快慢结点相遇的 点的距离是B
A+B+N = 2*(A+B)
A+B = N
所以 快慢指针相遇后，从头结点开始再跑一个慢指针，直到2个慢的相遇，相遇的点就是cycle begin
1 public class Solution { 2 public ListNode detectCycle(ListNode head) { 3 ListNode faster = head; 4 ListNode slower = head; 5 6 while(faster !=null && faster.next != null){ 7 faster = faster.next.next; 8 slower = slower.next; 9 10 if(faster == slower){ 11 ListNode slower2 = head; 12 while(slower != slower2){ 13 slower = slower.next; 14 slower2 = slower2.next; 15 } 16 return slower; 17 } 18 } 19 return null; 20 21 } 22 }
标签：bsp head cycle col can lower nod node return
原文地址：http://www.cnblogs.com/zle1992/p/7679160.html
