PAT-1074 Reversing Linked List

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博客围绕链表每K项翻转问题展开，给出输入输出规范及示例。提到改指针做法易出错，考试用STL容器更简单，但考点是指针，还给出直接改指针的版本及偷懒做法。

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1074 Reversing Linked List（25 分）

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

将链表每K项翻转，改指针的做法容易出错，不好检查，考试时用STL容器倒腾一下最简单。考点肯定是指针，以下是直接改指针的版本。

#include<stdio.h>

typedef struct node{
	int data,next;
}node; 
node l[100005];
int main(){
	int head,n,K;
	scanf("%d%d%d",&head,&n,&K);
	for(int i=0;i<n;i++){
		int a,v,n;
		scanf("%d %d %d",&a,&v,&n);
		l[a].data=v;l[a].next=n;
	}
	int size=0;
	for(int i=head;i!=-1;i=l[i].next){
		size++;
	}
	int count=size/K;
	int lastr=-1,th=head;
	for(int i=0;i<count;i++){
		int j=th,jj=l[j].next;
		for(int k=0;k<K-1;k++){
			int jjj=l[jj].next;
			l[jj].next=j;
			j=jj;jj=jjj;
		}
		l[th].next=jj;
		if(lastr!=-1)l[lastr].next=j;
		else head=j;
		lastr=th;
		th=jj;
	}
	int t=head;
	for(;l[t].next!=-1;t=l[t].next){
		printf("%05d %d %05d\n",t,l[t].data,l[t].next);
	}
	printf("%05d %d %d\n",t,l[t].data,l[t].next);

	return 0;
}

以下是偷懒的做法

#include<stdio.h>
#include<vector>
#include<algorithm>
using namespace std;

typedef struct node{
	int addr,data,next;
}node; 
node l[100005];
int main(){
	int head,n,K;
	scanf("%d%d%d",&head,&n,&K);
	for(int i=0;i<n;i++){
		int a,v,n;
		scanf("%d %d %d",&a,&v,&n);
		l[a].addr=a;l[a].data=v;l[a].next=n;
	}
	vector<node> ans;
	int size=0;
	for(int i=head;i!=-1;i=l[i].next){
		ans.push_back(l[i]);
		size++;
	}
	int count=size/K;
	int tmp=head;
	for(int i=0;i<count;i++){
		reverse(ans.begin()+i*K,ans.begin()+i*K+K);
	}
	int i=0;
	for(;i<ans.size()-1;i++){
		printf("%05d %d %05d\n",ans[i].addr,ans[i].data,ans[i+1].addr);
	}
	printf("%05d %d -1\n",ans[i].addr,ans[i].data);

	return 0;
}