lintcode102- Linked List Cycle- medium

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

Example

Given -21->10->4->5, tail connects to node index 1, return true

Challenge 

Follow up:
Can you solve it without using extra space?

public boolean hasCycle(ListNode head) 

1.O(n)空间。用HashSet，遍历链表的过程中，把结点加入到set里，如果遇到了碰到过的结点就说明有环。

2.O(1)空间。用快慢两指针。一个速度1(crt.next)，一个速度2(crt.next.next)，如果两者最后相遇说明一定是有环的。

1.HashSet实现

/**
 * Definition for ListNode.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int val) {
 *         this.val = val;
 *         this.next = null;
 *     }
 * }
 */


public class Solution {
    /*
     * @param head: The first node of linked list.
     * @return: True if it has a cycle, or false
     */
    public boolean hasCycle(ListNode head) {
        // write your code here
        Set<ListNode> set = new HashSet<>();
        while (head != null) {
            if (set.contains(head)) {
                return true;
            }
            set.add(head);
            head = head.next;
        }
        return false;
    }
}

2.快慢指针实现

/**
 * Definition for ListNode.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int val) {
 *         this.val = val;
 *         this.next = null;
 *     }
 * }
 */


public class Solution {
    /*
     * @param head: The first node of linked list.
     * @return: True if it has a cycle, or false
     */
    public boolean hasCycle(ListNode head) {
        // write your code here
        
        ListNode slow = head;
        ListNode quick = head;
        // 注意这个巧妙精简的写法，一旦出现null就肯定无环，正好借助它避免空指针
        while (quick != null && quick.next != null && slow != null) {
            slow = slow.next;
            quick = quick.next.next;
            if (slow == quick) {
                return true;
            }
        }
        return false;
    }
}