1074 Reversing Linked List

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 (≤) 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.

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

/*
    Name:
    Copyright:
    Author:  流照君
    Date: 2019/8/5 14:52:14
    Description:
*/
#include <iostream>
#include<string>
#include <algorithm>
#include <vector>
using namespace std;
const int inf=100100;
int main(int argc, char** argv)
{
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    int num=0,first,n,k,list[inf],add[inf],temp,next[inf];
    cin>>first>>n>>k;
    for(int i=1;i<=n;i++)
    {
    	cin>>temp;
    	cin>>add[temp]>>next[temp];
	}
	while(first!=-1)  //不是所有输入都是有用的 
	{
		list[num++]=first;
		first=next[first];
	}
	for(int i=0;i<num-num%k;i=i+k)
	reverse(begin(list)+i,begin(list)+i+k); //反转函数 
	for(int i=0;i<num-1;i++)
	printf("%05d %d %05d
",list[i],add[list[i]],list[i+1]);
	printf("%05d %d -1", list[num - 1], add[list[num - 1]]);
    return 0;
}

　　特别说一下

C++11引入了 begin 和 end 的函数，这两个函数与容器中的两个同名成员功能类似，不过这两个函数不是成员函数，而是含有参数的函数。

用法说明：

begin 返回首元素的地址，end 返回尾元素的下一个地址。