第10题 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?

Solution:

/** * Definition for singly-linked list. * class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { *         val = x; *         next = null; *     } * } */ public class Solution {    public ListNode detectCycle(ListNode head) {        if(head==null) return null;        ListNode ptr1=head, ptr2=head;       while(ptr1!=null && ptr2!=null){         if(ptr1.next==null || ptr2.next==null || ptr2.next.next==null) return null;         ptr1=ptr1.next;         ptr2=ptr2.next.next;         if(ptr1==ptr2)            break;       }       ptr1=head;       while(ptr1!=ptr2){           ptr1=ptr1.next;           ptr2=ptr2.next;       }       return ptr1;            }}

设圈前长度为x, 两指针在圈中相遇位置为k,圈周长为r。设相遇时ptr1跑了m圈，ptr2跑了n圈。因为ptr2速度是ptr1两倍，有：

2(x+k+mr) = x+k+nr

x+k = (n-m)r

因n和m均为整数且n>m，x+k应为整数圈。因为ptr1和ptr2在圈内位置为k，再跑x会到达圈起始点。从head跑x，同样会到达圈的起始点。所以讲ptr1移动到head，和ptr2以步长为1一起移动，相遇处就是圈的起始点。