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?
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/** * 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) { ListNode entry = null; boolean
flag = true; if(head != null){ ListNode speedOne = head; ListNode speedTwo = head; if(head.next != null){ speedOne = speedOne.next; if(speedOne.next != null) speedTwo = speedOne.next; else flag = false; } else flag = false; while(flag){ if(speedOne == speedTwo) break; else{ speedOne = speedOne.next; if(speedTwo != null
&& speedTwo.next != null){ speedTwo = speedTwo.next; speedTwo = speedTwo.next; } else{ flag = false; break; } } } if(flag){ speedOne = head; while(speedOne != speedTwo){ speedOne = speedOne.next; speedTwo = speedTwo.next; } entry = speedOne; } else entry = null; } return
entry; }} |
leetcode--Linked List Cycle II
原文：http://www.cnblogs.com/averillzheng/p/3545283.html