1074. Reversing Linked List (25)

题目链接：http://www.patest.cn/contests/pat-a-practise/1074

题目：

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

分析：

调换顺序，不过和地址有关，建立结构体。

注意点：节点顺序变换后地址也需要变换。

AC代码：

#include<stdio.h>
#include<iostream>
#include<vector>
using namespace std;
const int MAX = 100002;
struct Node{
 int add;
 int data;
 int next;
 Node(){ next = -1; }
}buf[MAX];
vector<Node>V_input;
vector<Node>V_ans;
int main(void){
 freopen("F://Temp/input.txt", "r",stdin);
 int start_add, all_nodes, N;
 scanf("%d%d%d", &start_add, &all_nodes, &N);
 for (int i = 0; i < all_nodes; i++){
  int cur_add, next_add, data;
  scanf("%d%d%d", &cur_add, &data,&next_add);
  buf[cur_add].add = cur_add;
  buf[cur_add].next = next_add;
  buf[cur_add].data = data;
 }
 int i;
 for (i = start_add; buf[i].next != -1; i = buf[i].next){
  V_input.push_back(buf[i]);
 }
 V_input.push_back(buf[i]);
 int idx = 0;
 while (idx + N <= V_input.size()){
  //这里要用V_input.size()而不是用all_nodes，因为会有无用节点的存在
  for (int i = idx + N - 1; i >= idx; i--){
   Node *tmp = &V_input[i];
   if (i != idx){
    tmp->next = V_input[i - 1].add;
   }//从后向前来，next地址变成前一个节点的地址
   else if(idx + N != V_input.size()){
    tmp->next = V_input[idx + N].add;
   }//第一个节点的地址变成向后第N个节点的地址
   else {
    tmp->next = -1;
   }//当碰到10个节点要55倒序的时候，要指定为-1，而不能找到向后第N个节点的地址
   V_ans.push_back(*tmp);
  }
  idx += N;
 }
 for (int i = idx; i < V_input.size(); i++){
  V_ans.push_back(V_input[i]);
 }
 for (int i = 0; i < V_input.size(); i++){
  if (i == V_input.size() - 1){
   printf("%05d %d -1\n", V_ans[i].add, V_ans[i].data);
  }
  else{
   printf("%05d %d %05d\n", V_ans[i].add, V_ans[i].data, V_ans[i+1].add);
  }
 }
 return 0;
}

截图：

——Apie陈小旭