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
简单的模拟题,以空间换时间：

#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;const int NUM=100001;struct Node
{int address;int data;int next;
};Node nodes[NUM];
vector<Node> list;
int main()
{int fnAddress,N,K;scanf("%d %d %d",&fnAddress,&N,&K);int i;for(i=0;i<N;++i){Node node;scanf("%d %d %d",&node.address,&node.data,&node.next);nodes[node.address]=node;}int address=fnAddress;while(address!=-1)//去噪
    {list.push_back(nodes[address]);address=nodes[address].next;}int size=list.size();int round=size/K;int start,end;for(i=1;i<=round;++i){start=(i-1)*K;end=i*K;reverse(list.begin()+start,list.begin()+end);}for(i=0;i<size-1;++i){printf("%.5d %d %.5d\n",list[i].address,list[i].data,list[i+1].address);}printf("%.5d %d %d\n",list[i].address,list[i].data,-1);return 0;
}

View Code

转载于:https://www.cnblogs.com/wwblog/p/3708838.html

PAT 1074. Reversing Linked List (25)相关推荐

PAT 1074 Reversing Linked List
1074 Reversing Linked List (25point(s)) Given a constant K and a singly linked list L, you are suppo ...
1074. Reversing Linked List (25)-PAT甲级真题
Given a constant K and a singly linked list L, you are supposed to reverse the links of every K elem ...
1074 Reversing Linked List (25 分) java 题解
Given a constant K and a singly linked list L, you are supposed to reverse the links of every K elem ...
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 elem ...
1074 Reversing Linked List (25 分)【难度: 一般 / 知识点: 链表】
https://pintia.cn/problem-sets/994805342720868352/problems/994805394512134144 乙级里面的原题吧,就用哈希表建链表,reve ...
PAT 1074 Reversing Linked List——双端队列解法
题目地址原始想法既然给定的序列是乱序的: 但是输出要求顺序,再按块反转输出那么我为何不直接在读入的时候调整为顺序呢! 具体思路读入1个结点后,剩下的结点,仅存在以下3种情况插在他前面:插在他 ...
PAT甲级1074 Reversing Linked List ：[C++题解]反转链表，借用vector
文章目录题目分析题目链接题目分析分析:数组模拟链表,这题反转操作在数组中进行,然后直接输出即可,甚至不用放回到链表. //遍历链表,该链表用数组模拟 //保存链表结点地址到数组中 for(in ...
每隔k次反转一次链表_PTA 5-2 Reversing Linked List (25) [法一] - 线性表 - 链表反转 (PAT 1074)...
Given a constant K and a singly linked list L, you are supposed to reverse the links of every K elem ...
PAT-A 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 elem ...

PAT 1074. Reversing Linked List (25)

PAT 1074. Reversing Linked List (25)相关推荐

最新文章

热门文章