P14-环形链表-快慢指针

2021-05-13  本文已影响0人  YonchanLew
//环形链表
/*
* 给定一个链表,判断链表中是否有环
* 如果链表中有某个节点,可以通过连续跟踪next指针再次到达该节点,则链表中存在环
* 如果链表中存在环,返回true,否则false
* */
public class P14 {

    static class ListNode{
        int val;
        ListNode next;

        public ListNode(int val, ListNode next){
            this.val = val;
            this.next = next;
        }
    }

    public static void main(String[] args) {
        ListNode node5 = new ListNode(5, null);
        ListNode node4 = new ListNode(4, node5);
        ListNode node3 = new ListNode(3, node4);
        ListNode node2 = new ListNode(2, node3);
        ListNode node1 = new ListNode(1, node2);

        node5.next = node3;

        System.out.println(hasCycle(node1));
        System.out.println(hasCycle2(node1));

    }

    private static boolean hasCycle(ListNode head) {

        //空间 O(n)
        Set<ListNode> set = new HashSet<>();

        //时间 O(n)
        while(head != null){
            //添加失败会返回false
            if(!set.add(head)){
                return true;
            }
            head = head.next;
        }

        return false;
    }

    //双指针,快慢指针,龟兔赛跑
    private static boolean hasCycle2(ListNode head) {

        if(head == null || head.next == null){
            return false;
        }

        ListNode slow = head;
        ListNode quick = head.next;

        //时间 O(n)
        while(slow != quick){
            if(quick == null || quick.next == null){
                return false;
            }
            slow = slow.next;
            //走得快,同时减少遍历次数
            //快指针提前进入环,迟早和慢指针相遇
            quick = quick.next.next;
        }

        return true;
    }
}
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