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

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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 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

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;
}

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