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中国大学MOOC-陈越、何钦铭-数据结构-2018秋 02-线性结构3 Reversing Linked List (25 分)
阅读量:3904 次
发布时间:2019-05-23

本文共 1389 字,大约阅读时间需要 4 分钟。

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 (≤10​5​​) 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

代码如下:

#include 
#include
#include
#include
#include
using namespace std;const int maxn=1e6+5;int st,n,re;struct List{ int data; int next;};int LL[maxn],num=0;List L[maxn];int main(){ scanf("%d%d%d",&st,&n,&re); for (int i=0;i

 

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