【PAT】1074. Reversing Linked List (25)

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

题意：给出链表，每k个元素进行反转，输出反转后的链表。
分析：首先根据input模拟构建一个链表，再将链表按顺序存入一个vector，然后用reverse函数将vector中的元素每k个进行反转就行了。

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
#define MAX 1000000

struct node{
	int addr;
	int next;
	int val;
};

int main(int argc, char** argv) {
	int begin, n, k;
	scanf("%d%d%d",&begin, &n, &k); 
	int i;
	vector<node> vec(MAX);
	int addr, next, val;
	for(i=0; i<n; i++){
	 	scanf("%d%d%d", &addr,&val,&next);
	 	vec[addr].addr = addr;
	 	vec[addr].next = next;
	 	vec[addr].val = val;
	}
	int cnt=0, j;
	
	vector<node> result;
	for(i=begin; i!=-1; i=vec[i].next){
		result.push_back(vec[i]);		
	}
	
	int from , to;
	for(i=0; i<result.size(); i++){
		from = i*k;
		to = from + k;
		if(to > result.size()){
			break;
		}else{
			reverse(result.begin()+from, result.begin()+to);
		}
	}
	//cout<<"result: "<<endl;
	if(result.size() == 1){
		printf("%05d %d -1\n",result[0].addr, result[0].val);
		return 0;
	}
	
	for(i=0; i<result.size(); i++){
		if(i == 0){
			printf("%05d %d", result[i].addr, result[i].val);
		}else if(i==result.size()-1){
			printf(" %05d\n%05d %d -1\n", result[i].addr,result[i].addr,result[i].val);			
		}else{
			printf(" %05d\n%05d %d", result[i].addr,result[i].addr,result[i].val );
		}
	}	
 
	return 0;
}