zoukankan      html  css  js  c++  java
  • LeetCode Reverse Nodes in k-Group

    Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.

    If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.

    You may not alter the values in the nodes, only nodes itself may be changed.

    Only constant memory is allowed.

    For example,
    Given this linked list: 1->2->3->4->5

    For k = 2, you should return: 2->1->4->3->5

    For k = 3, you should return: 3->2->1->4->5

    沿用的另一道题目的代码,写的不太好,有空了需要重新写过。

      1 /**
      2  * Definition for singly-linked list.
      3  * public class ListNode {
      4  *     int val;
      5  *     ListNode next;
      6  *     ListNode(int x) {
      7  *         val = x;
      8  *         next = null;
      9  *     }
     10  * }
     11  */
     12 public class Solution {
     13     public ListNode reverseKGroup(ListNode head, int k) {
     14 
     15         if (head==null) {
     16             return null;
     17         }
     18         if (k == 1) {
     19             return head;
     20         }
     21 
     22         ListNode temp=head;
     23         ListNode doRe = head, nextList = null, first = null, last = null;
     24         int i=1;
     25 
     26         while (temp != null) {
     27 
     28             if (i % k == 0) {
     29                 if (i == k) {
     30                     first = reverseBetween(doRe, i / k, i);
     31 
     32                 } else {
     33                     last.next = reverseBetween(doRe, 1, k);
     34                 }
     35                 last = doRe;
     36                 temp = doRe.next;
     37                 doRe = doRe.next;
     38             } else {
     39             //    last = temp;
     40                 temp = temp.next;
     41             }
     42 
     43             ++i;
     44         }
     45         return first==null?doRe:first;
     46 
     47     }
     48     
     49     
     50     public ListNode reverseBetween(ListNode head, int m, int n) {
     51         if (head==null) {
     52             return null;
     53         }
     54         if(n==m){
     55             return head;
     56         }
     57         ListNode temp=head,doRe = null,second=null,first=null,last=null;
     58         int i=1;
     59 
     60         while (temp!=null) {
     61             if (i==m-1) {
     62                 first=temp;
     63             }
     64             if (i==m) {
     65                 doRe=temp;
     66             }
     67             if (i==n) {
     68                 last=temp;
     69                 second=temp.next;
     70                 if(first!=null) first.next=last;
     71                 last.next=null;
     72 
     73                 break;
     74             }
     75             ++i;
     76             temp=temp.next;
     77         }
     78         reverse(doRe);
     79         doRe.next=second;    
     80 
     81         if (m>1) {
     82             return head;
     83         }else {
     84             return last;
     85         }
     86 
     87     }
     88     private void reverse(ListNode head) {
     89         ListNode pre = null,curr,next;
     90 
     91         curr=head;
     92         if (head==null) {
     93             return;
     94         }
     95         while (curr!=null) {
     96             next=curr.next;
     97             curr.next=pre;
     98             pre=curr;
     99             curr=next;
    100         }
    101     }
    102 
    103 
    104 
    105 }
  • 相关阅读:
    ExtJS4学习笔记二--表单控件相关
    Js中replace()的用法
    浅析轮询(Polling)和推送(LongPolling)服务
    ExtJS4学习笔记五--面板使用
    ExtJS4学习笔记四--图片上传
    spring MVC
    ExtJS4学习笔记三--远程访问数据源示例
    Struts 2
    ExtJs4学习笔记一--基础知识
    URL编码规则
  • 原文地址:https://www.cnblogs.com/birdhack/p/4174950.html
Copyright © 2011-2022 走看看