Linked List Cycle

Given a linked list, determine if it has a cycle in it.

Follow up:

Can you solve it without using extra space?

Floyd判圈算法的典型应用场景

①使用Floyd判圈算法直接求解

/** * Definition for singly-linked list. * class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { *         val = x; *         next = null; *     } * } */public class Solution {    public boolean hasCycle(ListNode head) {        if (head == null || head.next == null) {            return false;        }        ListNode slow = head;        ListNode fast = head;                while (fast != null && fast.next != null) {            slow = slow.next;            fast = fast.next.next;            if (slow == fast) {                return true;            }        }        return false;            }}

程序中需要注意的是while循环的终止条件是(fast != null && fast.next != null)，缺一不可， 否则会有问题。

②逆转链表，判断前后两个链表的头结点是否相同。如果链表有环，则逆转后的链表和初始链表的头结点是相同的。

/** * Definition for singly-linked list. * class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { *         val = x; *         next = null; *     } * } */public class Solution {    public boolean hasCycle(ListNode head) {        if (head == null || head.next == null) {            return false;        }        if (head == reverseList(head)) {            return true;        }        return false;    }        private ListNode reverseList(ListNode head) {        ListNode prev = null;        ListNode next = null;                while (head != null) {            next = head.next;            head.next = prev;            prev = head;            head = next;        }                return prev;    }}