leetcode 142. Linked List Cycle II ----- java

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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?

 

141題的延伸，求出迴圈點。

 

 

可以用數學方法證明出slow與find相遇的位置一定是所求的點。

 

/** * 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 || head.next == null )            return null;        ListNode fast = head;        ListNode slow = head;        while( fast != null && fast.next != null  ){            slow = slow.next;            fast = fast.next.next;            if( fast == slow ){                ListNode find = head;                while( find != slow ){                    find = find.next;                    slow = slow.next;                }                return find;            }        }        return null;            }}

 

leetcode 142. Linked List Cycle II ----- java