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Total Accepted: 112637
Total Submissions: 363331
Difficulty: Medium
Contributor: LeetCode
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?
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Linked List Two Pointers
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(E) Linked List Cycle (M) Find the Duplicate Number
** 解题思路 **
用快慢指针，快指针一次走两步，慢指针一次走一步，来先确定是否有cycle。
如果存在cycle，则将fast = head，再重新 fast =fast.next, slow = slow.next 两个指针都单步走，找到fast = slow点即可确定cycle起始点。
/**
* 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 slow = head;
ListNode fast = head;
boolean isCycle = false;
while (fast != null && fast.next != null) {
fast = fast.next.next;
slow = slow.next;
if (fast == slow) {
isCycle = true;
break; // don't forget to break the loop
}
}
if (!isCycle) return null;
// find the node while cycle begins
fast = head;
while (fast != slow) {
fast = fast.next;
slow = slow.next;
}
return slow;
}
}
