Reversing Linked List

Posted 2020-10-28 又啦

Reversing Linked List

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 (≤) which is the total number of nodes, and a positive K (≤) 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

AC Code

 1 #include<stdio.h>
 2 typedef struct _Node{              //创建节点
 3   int data; 
 4   int next;
 5 }node;
 6 void Input_array(node a[],int N);
 7 void reverse_once(node a[],int *sub_start,int k);
 8 void Output_array(node a[],int start);
 9 int main(){
10   node a[100002];          //节点数组
11   int N;
12   int k;
13   int times_of_reverse;     //需要反转的次数
14   int sub_start;            //子头节点
15   int n;
16   int p;
17   //读入
18   scanf("%d %d %d",&a[100001].next,&N,&k);
19   Input_array(a,N);
20   //处理
21   p = a[100001].next;
22   n = 0;
23   while(p!=-1){           //找到有效的个数
24     n++;
25     p = a[p].next;
26   }
27   times_of_reverse = n/k;
28   sub_start = 100001;           //头节点
29   while(times_of_reverse--){
30     reverse_once(a,&sub_start,k);
31   }
32   
33   //输出
34   Output_array(a,a[100001].next);
35   return 0;
36 }
37 
38 void Input_array(node a[],int N){                       //读入
39   int addr_tem,data_tem,next_tem;
40   while(N--){
41     scanf("%d %d %d",&addr_tem,&data_tem,&next_tem);
42     a[addr_tem].data = data_tem;
43     a[addr_tem].next = next_tem;
44   }
45 }
46 void reverse_once(node a[],int *sub_start,int k){          //反转一次
47   int thisP,preP,lastP,temP;     //比经典单链表的反转多记录一个需返回的下一个头节点，temP
48   thisP = a[*sub_start].next;
49   temP = thisP;
50   lastP = -1;
51   for(int i = 0; i < k ; i++){
52     preP = a[thisP].next;
53     a[thisP].next = lastP;
54     lastP = thisP;
55     thisP = preP;
56   }
57   a[temP].next = thisP;        //普通反转没有这一步
58   a[*sub_start].next = lastP;
59   *sub_start = temP;
60 }
61 void Output_array(node a[],int start){
62   int p = start;
63   while(a[p].next!= -1){
64     printf("%05d %d %05d\\n",p,a[p].data,a[p].next);
65     p = a[p].next;
66   }
67   printf("%05d %d %d\\n",p,a[p].data,a[p].next);
68   
69 }