博客
关于我
牛客网算法——名企高频面试题143题(13)
阅读量:400 次
发布时间:2019-03-04

本文共 2610 字,大约阅读时间需要 8 分钟。

????

?????????????????????????????????????????????????????????????

???????

??

?????????????????????????????

  • ????????pre?curr??????????????????
  • ??????????????????
  • ?????????????????pre???????
  • ??pre?curr??????????????????
  • ???????pre????????????
  • ???????????O(n)???????O(1)???????????????

    ????

    public class ????II {    public class ListNode {        int val;        ListNode next;        public ListNode(int val) {            this.val = val;        }    }    public ListNode rever(ListNode head) {        if (head == null) {            return null;        }        ListNode pre = null;        ListNode curr = head;        while (curr != null) {            ListNode future = curr.next;            curr.next = pre;            pre = curr;            curr = future;        }        return pre;    }        public static void main(String[] args) {        // ????        ListNode s1 = new ListNode(1);        ListNode s2 = new ListNode(2);        ListNode s3 = new ListNode(3);        ListNode s4 = new ListNode(4);        s1.next = s2;        s2.next = s3;        s3.next = s4;        ListNode res = rever(s1);        while (res != null) {            System.out.println(res.val);            res = res.next;        }    }}

    ??????????

    ????????????????????????????????????????????

  • ????dummy??????????????
  • ???????????dummy?????????
  • ????????pre?curr?????dummy?????????
  • ????????????????????
  • ?????????dummy????????????????????
  • ????????O(n)???????O(1)???????

    ????

    public class ????II {    public class ListNode {        int val;        ListNode next;        public ListNode(int val) {            this.val = val;        }    }    public ListNode rever1(ListNode head) {        if (head == null) {            return null;        }        ListNode dumy = new ListNode(0);        dumy.next = head;        ListNode pre = dumy;        ListNode curr = head;        while (curr.next != null) {            ListNode future = curr.next;            curr.next = future.next;            future.next = dumy.next;            pre.next = future;        }        return dumy.next;    }        public static void main(String[] args) {        // ????        ListNode s1 = new ListNode(1);        ListNode s2 = new ListNode(2);        ListNode s3 = new ListNode(3);        ListNode s4 = new ListNode(4);        s1.next = s2;        s2.next = s3;        s3.next = s4;        ListNode res = rever1(s1);        while (res != null) {            System.out.println(res.val);            res = res.next;        }    }}

    ????

    ???????1?2?3?4???????4?3?2?1?????????????????

    4321

    ??

    ???????????????????????????????????????????????????????????????????????????????

    转载地址:http://wqch.baihongyu.com/

    你可能感兴趣的文章
    Palindrome Number leetcode java
    查看>>
    Palo Alto Networks Expedition 未授权SQL注入漏洞复现(CVE-2024-9465)
    查看>>
    Palo Alto Networks Expedition 远程命令执行漏洞(CVE-2024-9463)
    查看>>
    Palo Alto Networks PAN-OS身份认证绕过导致RCE漏洞复现(CVE-2024-0012)
    查看>>
    Panalog 日志审计系统 libres_syn_delete.php 前台RCE漏洞复现
    查看>>
    Springboot中@SuppressWarnings注解详细解析
    查看>>
    Panalog 日志审计系统 sprog_deletevent.php SQL 注入漏洞复现
    查看>>
    Panalog 日志审计系统 sprog_upstatus.php SQL 注入漏洞复现(XVE-2024-5232)
    查看>>
    Panalog 日志审计系统 前台RCE漏洞复现
    查看>>
    PANDA VALUE_COUNTS包含GROUP BY之前的所有值
    查看>>
    Pandas - 有条件的删除重复项
    查看>>
    pandas -按连续日期时间段分组
    查看>>
    pandas -更改重新采样的时间序列的开始和结束日期
    查看>>
    pandas :to_excel() float_format
    查看>>
    pandas :加入有条件的数据框
    查看>>
    pandas :将多列汇总为一列,没有最后一列
    查看>>
    pandas :将时间戳转换为 datetime.date
    查看>>
    pandas :将行取消堆叠到新列中
    查看>>
    pandas DataFrame 中的自定义浮点格式
    查看>>
    Pandas DataFrame 的 describe()方法详解-ChatGPT4o作答
    查看>>