首页 > 其他 > 详细

在时间复杂度为O(n)且空间复杂度为O(1)的情况下翻转链表

时间:2020-06-20 16:40:52      阅读:84      评论:0      收藏:0      [点我收藏+]
class NodeList{
    private Node head;
    private Node tail;

    void print(){
        Node node = head;
        while (node != null){
            System.out.print(node.date+",");
            node = node.next;
        }
        System.out.println();
    }
    private void add(int val){
        Node newNode = new Node(val);
        if(head == null){
            head = newNode;
            tail = newNode;
        }else {
            tail.next = newNode;
            tail = newNode;
        }
    }

    private void rever(){
        tail = head;
        Node pre = null;
        Node temp = null;
        Node p = head;
        while (p != null){
            temp = p.next;
            p.next = pre;
            pre = p;
            p = temp;
        }
        head = pre;
    }

    public static void main(String[] args) {
        NodeList list = new NodeList();
        list.add(1);
        list.add(2);
        list.add(3);

        list.rever();
        list.print();
    }
}

class Node{
    public int date;
    public Node next;



    public Node(int date) {
        this.date = date;
    }
}

 

在时间复杂度为O(n)且空间复杂度为O(1)的情况下翻转链表

原文:https://www.cnblogs.com/dongma/p/13169041.html

(0)
(0)
   
举报
评论 一句话评论(0
关于我们 - 联系我们 - 留言反馈 - 联系我们:wmxa8@hotmail.com
© 2014 bubuko.com 版权所有
打开技术之扣,分享程序人生!