标签：link flex ems repr blank 反转 cout ack ica
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.
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.
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.
00100 6 4
00000 4 99999
00100 1 12309
68237 6 -1
33218 3 00000
99999 5 68237
12309 2 33218
00000 4 33218 33218 3 12309 12309 2 00100 00100 1 99999 99999 5 68237 68237 6 -1
天哪！一个链表写一天
#include <iostream> #include <stdio.h> using namespace std; struct Node { int Num; int Next; //下一个地址 }List[100001]; //地址100000作为开头 void print(int n) { if(n==-1) { cout<<-1; return ; } int t=10000; while(t) { cout<<n/t; n=n%t; t=t/10; } } int main() { int fir,N,K; scanf("%d %d %d",&fir,&N,&K); //cin>>fir>>N>>K; List[100000].Next=fir; for(int i=0;i<N;++i) { int t1,t,t2; //cin>>t1>>t>>t2; scanf("%d %d %d",&t1,&t,&t2); List[t1].Num=t; List[t1].Next=t2; } int P1=List[fir].Next,P2=List[fir].Next,P3=fir,P4=100000; while(1) {//cout<<"P4="<<P4<<" P3="<<P3<<" P2="<<P2<<" P1="<<P1<<endl; int flag=1; for(int i=1;i<K;i++) { if(P1!=-1) { P1=List[P1].Next; List[P2].Next=P3; P3=P2; P2=P1; } else { flag=0; break; } //cout<<"aaP4="<<P4<<" P3="<<P3<<" P2="<<P2<<" P1="<<P1<<endl; } if(flag) { int y=List[P4].Next; List[P4].Next=P3; P3=y; //cout<<"P3="<<P3<<endl; //for(int j=1;j<K;++j) P3=List[P3].Next; List[P3].Next=P1; //复原 P4=P3; P3=P2; if(P2==-1) break; //刚好结束 P1=List[P1].Next; P2=List[P2].Next; //cout<<"P4="<<P4<<" P3="<<P3<<" P2="<<P2<<" P1="<<P1<<endl; } else//结束，有尾巴,撤回操作(debug:多退了一步，啊啊啊啊阿) { while(P3!=List[P4].Next)//!!! { //P1=P2; P2=P3; P3=List[P3].Next; List[P2].Next=P1; P1=P2; } break; } } //cout<<endl; int P=100000; while(1) { P=List[P].Next; print(P); cout<<" "<<List[P].Num<<" "; print(List[P].Next); cout<<endl; if(List[P].Next==-1) break; } return 0; }
02-线性结构3 Reversing Linked List (25分)(链表每段反转)
标签：link flex ems repr blank 反转 cout ack ica
原文地址：https://www.cnblogs.com/liuyongliu/p/12466163.html
