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, you must output 4→3→2→1→5→6.
Input Specification:
Each input file contains one test case. For each case, the first line contains the address of the first node, a positive N (<= 105) which is the total number of nodes, and a positive K (<=N) which is the length of the sublist to be reversed. The address of a node is a 5-digit nonnegative integer, and NULL is represented by -1.
Then N lines follow, each describes a node in the format:
Address Data Next
where Address is the position of the node, Data is an integer, and Next is the position of the next node.
Output Specification:
For each case, output the resulting ordered linked list. Each node occupies a line, and is printed in the same format as in the input.
Sample Input:
00100 6 4 00000 4 99999 00100 1 12309 68237 6 -1 33218 3 00000 99999 5 68237 12309 2 33218
Sample Output:
00000 4 33218 33218 3 12309 12309 2 00100 00100 1 99999 99999 5 68237 68237 6 -1 思路：这道题我花了好长时间去解析，为什么花了好长时间， 首先没有读懂题，其次在没读懂题的基础上纠结了好长时间就是临时变量的问题。最后再读懂题的基础上没有考虑边界的情况。总之：受益匪浅 1：这个是错误的代码： 错误的代码必须一辈子铭记。
1 #include <iostream>
2 #include <cstdio>
3 using namespace std;
4 #define MAX 100010
5 struct Info
6 {
7 int pre;
8 int data;
9 int next;
10 }node[MAX];
11 //一定要仔细读题....every k elements
12 int main(int argc, char *argv[])
13 {
14 int start,N,K;
15 scanf("%d%d%d",&start,&N,&K);
16 for(int i=0;i<N;i++)
17 {
18 int pre,data,next;
19 scanf("%d%d%d",&pre,&data,&next);
20 node[pre].data=data;
21 node[pre].next=next;
22 }
23 int pt=start;
24 //进行前驱记录
25 node[pt].pre=-2;//开头的前驱代表着开始
26 while(node[pt].next!=-1)
27 {
28 int temp=pt;
29 pt=node[pt].next;
30 node[pt].pre=temp;
31 }
32 //寻找第K个
33 pt=start;
34 for(int i=1;i<K&&node[pt].next!=-1;i++)
35 {
36 pt=node[pt].next;
37 }
38 //进行反转
39 int temp=node[pt].next;
40 start=pt;
41 while(node[pt].pre!=-2)
42 {
43 node[pt].next=node[pt].pre;
44 pt=node[pt].pre;
45 }
46 node[pt].next=temp;
47 //进行结果的输出
48 pt=start;
49 while(node[pt].next!=-1)
50 {
51 printf("%05d %d %05d\n",pt,node[pt].data,node[pt].next);
52 pt=node[pt].next;
53 }
54 printf("%05d %d %d\n",pt,node[pt].data,node[pt].next);
55 return 0;
2:正确的代码
1 #include <cstdio>
2 #include <vector>
3 using namespace std;
4 #define MAX 100010
5 struct node
6 {
7 int add;
8 int data;
9 int next;
10 };
11 int main()
12 {
13 vector<node>in(MAX); //像数组一样
14 vector<node>srt;
15 vector<node>out;
16 int start,N,K;
17 scanf("%d%d%d",&start,&N,&K);
18 for(int i=0;i<N;i++)
19 {
20 node temp;
21 scanf("%d%d%d",&temp.add,&temp.data,&temp.next);
22 in[temp.add]=temp;
23 }
24 int next=start;
25 while(next!=-1)
26 {
27 srt.push_back(in[next]);
28 next=in[next].next;
29 }
30 int left=0;
31 while(left<srt.size())
32 {
33
34 for(int i=left+K-1;i>=left;i--)
35 {
36 if(i>=srt.size())
37 break;
38 out.push_back(srt[i]);
39 }
40 left+=K;
41 }
42 if(left!=srt.size())
43 left-=K;
44 while(left<srt.size())
45 {
46 out.push_back(srt[left]);
47 left++;
48 }
49 /*
50 int len=srt.size();
51 int right=K-1;
52 while(right<len)
53 {
54 for(int i=right;i>right-K;i--)
55 {
56 out.push_back(srt[i]);
57 }
58 right+=K;
59 }
60 right-=K;
61 right++;
62 for(int i=right;i<srt.size();i++)
63 {
64 out.push_back(srt[i]);
65 }*/
66 for(int i=0;i<out.size();i++)
67 {
68 if(i==out.size()-1)
69 printf("%05d %d -1\n",out[i].add,out[i].data);
70 else
71 printf("%05d %d %05d\n",out[i].add,out[i].data,out[i+1].add);
72 }
73 return 0;
74 }