seo 发表于 2022-5-31 13:33:26

1074. Reversing Linked List (25)

1074. Reversing Linked List (25)发布时间：2022/5/31 13:06:38
            
                                                       
                                                       
            
      
      
               
                     
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 (5) which is the total number of nodes, and a positive K (

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

#include
#include
using namespace std;
const int maxn = 100100;
struct Node{
    int address,data,next;
    int order;
}node;
bool cmp(Node a,Node b){
    return a.orderb.order;
}
int main(){
    int i,address;
    for(i = 0; i ){
      node.order = maxn;
    }
    int begin,n,k;//起始节点地址，节点数目，分组数
    scanf("%d%d%d",&begin,&n,&k);
   
    for(i = 0; i ){
      scanf("%d",&address);
      scanf("%d%d",&node.data,&node.next);
      node.address = address;
    }
    int p = begin,count = 0;
    while(p != -1){
      node.order = count++;
      p = node.next;
    }
    sort(node,node+maxn,cmp);
    n = count; //因为count=0占一个有效节点，退出循环时，count值就是有效节点
    for(i = 0; i //枚举完整的n/k块
       for(int j = (i+1)*k - 1; j > i*k; j-- ){ //每块的第i个倒着输出，剩余最后一个节点
         printf("%05d %d %05d\n",node.address,node.data,node.address);
       }
       printf("%05d %d ",node.address,node.data); //每块的最后一个节点的前两项数据
      if(i 1){ //如果是非最后一块节点
            printf("%05d\n",node[(i+2)*k-1].address);
      }else{   //如果是最后一块节点
            if(n % k == 0) printf("-1\n");//刚好除整
            else{   //如果最后一个节点不规则
            printf("%05d\n",node[(i+1)*k].address);
            for(i = n/k*k; i ){
                printf("%05d %d ",node.address,node.data);
                if(i 1){
                  printf("%05d\n",node.address);
                }else{
                  printf("-1\n");
               }// else
            } // for(i)
         }//else
      }//else
    } //for(i)
   
    return 0;
}




转载于:https://www.cnblogs.com/wanghao-boke/p/8550233.html
               
      
      
   
            
      
      
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