Given a constant K and a singly linked list L, you are supposed to reverse the links of every K elements on L. For example, given L being 1→2→3→4→5→6, if K = 3, then you must output 3→2→1→6→5→4; if K

题目 02-线性结构3 Reversing Linked List (25 分) Given a constant K and a singly linked list L, you are supposed to reverse the links of every K elements on L. For example, given L being 1→2→3→4→5→6, if K=3

1.下载附件，先查壳，无壳 2.用IDA分析，找到main函数，使用fn+f5，反编译 3.分析代码 4.双击进入条件函数中查看代码 5.编写代码，来源：https://blog.csdn.net/2303_80796023/article/details/137086655?ops_request_misc=%257B%2522request%255Fid%2522%25

题目描述 Given a constant K and a singly linked list L, you are supposed to reverse the links of every K elements on L. For example, given L being 1→2→3→4→5→6, if K = 3, then you must output 3→2→1→6→5→4;

Given a constant K and a singly linked list L, you are supposed to reverse the links of every K elements on L. For example, given L being 1→2→3→4→5→6, if K=3, then you must output 3→2→1→6→5→4; if K=4,

甲级传送门 乙级传送门 #include<stdio.h>#include<algorithm>using namespace std;#define MAX_N 100100struct Node{int address;int next;int data;int order;}node[MAX_N];bool cmp(struct Node a,struct Node b){retur

PAT 1074. Reversing Linked List，作为课后题出现在数据结构MOOC上。 刚开始学数据结构，水平还差的远 ，参考网上code，参考：http://tech-wonderland.net/blog/pat-1074-reversing-linked-list.html 路漫漫其修远兮 也是刚开始没有理解好题目意思，导致有几处感觉有难度，

文章目录 一、题目分析二、具体思路三、题目总结 Reverse is a cup of hot tea ,we need to drink it slowly. ：创造是极客唯一的属性 一、题目分析 一个枪战游戏，试玩发现场景内很多“潇洒哥”并且可以被清除。但是游戏体验不佳（20-30枪才能打死一个“潇洒哥”以及移动速度慢）。 IDA载入分析程序逻辑： 前面一大堆代码