Reversing Linked List (链表模拟

一般的链表套路,建结构体,然后从头开始找,把地址push进vector里
用reverse交换就可以了
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 (≤10
​5
​​ ) 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

#include
#include 
#include 
#include 
#include
#include
using namespace std;
struct nodee{
    int add;
    int val;
    int nxt;
}node[1000010];
 
int main() {
    int start,n,front;
    cin>>start>>n>>front;
    vectorv;
    for(int i=0;i>add;
        cin>>node[add].val>>node[add].nxt;
        node[add].add=add;
     } ;
     int t=start;
  
     
     while(node[t].nxt!=-1)
     { 
	 	v.push_back(node[t].add);
        t=node[t].nxt;
       
     }
      v.push_back(node[t].add);
      
    
    int cnt=0;
    while(cnt+front<=v.size())
    {
    	reverse(v.begin()+cnt,v.begin()+cnt+front);
		cnt+=front;
	}
//	cout<

你可能感兴趣的:(帕特)