02-1. Reversing Linked List (25) PAT DBS打基础之练习

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 400000 4 9999900100 1 1230968237 6 -133218 3 0000099999 5 6823712309 2 33218

Sample Output:

00000 4 3321833218 3 1230912309 2 0010000100 1 9999999999 5 6823768237 6 -1

思路比较明显:已知链表的首结点地址，我们只需要依次查找下一个结点，就可以得到链表的顺序，然后我们按照题目要求，每k个元素输出一次就行， 不足k个元素直接原序输出。 此题比较坑的是第6组测试数据，输入的节点中有无效节点。

#include<iostream>#include<cstdio>#include<cstdlib>#include<vector>#include<algorithm>#include<stack>#include<queue>#include<set>#include<map>#include<string>#include<cstring>using namespace std;const int mx=100005;typedef struct node{int add;int data;int nextAdd;}Node; Node nod[mx],ans[mx];int main(){int first,n,k;while(scanf("%d %d %d",&first,&n,&k)!=EOF){ int i=0,N=n;while(N--){scanf("%d",&i);nod[i].add=i;int t;scanf("%d",&t);nod[i].data=t;scanf("%d",&t);nod[i].nextAdd=t;}int j=0;for(int i=first;;){ans[j].add=nod[i].add;ans[j].data=nod[i].data;j++;i=nod[i].nextAdd;if(i == -1) break;}n=j;//处理无效结点。 int pre=0,stop=n/k*k-1;if(n%k==0) ans[n-1+k].add=-1;else ans[n-1].nextAdd=-1;//每k个结点逆序输出。 for(j=k-1;j<=stop;j+=k){int temp=j;while(temp > pre){ printf("%05d %d %05d\n",ans[temp].add,ans[temp].data,ans[temp-1].add);temp--;}pre=j+1;if(ans[j+k].add!=-1)if(n%k!=0 && j==stop) printf("%05d %d %05d\n",ans[temp].add,ans[temp].data,ans[j+1].add);else printf("%05d %d %05d\n",ans[temp].add,ans[temp].data,ans[j+k].add);else if(n%k==0)printf("%05d %d %d\n",ans[temp].add,ans[temp].data,ans[j+k].add);}if(n%k != 0){for(i=j+1-k;i<n;i++)if(ans[i].nextAdd!=-1)printf("%05d %d %05d\n",ans[i].add,ans[i].data,ans[i+1].add);else printf("%05d %d %d\n",ans[i].add,ans[i].data,ans[i].nextAdd);}   }return 0;}