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 (<= 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 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


题目大意：给出n个节点，以及头结点，要求把相邻的k个节点翻转， 如果n不能被k整除，剩下的节点按照原来的顺序
思路：先用vector<int> value保存当前节点的值，  vector<int> v保存当前节点指向下一个节点的指针； 然后再根据链表信息把k个链表的信息保存在vector<node> temp中， 然后在反向添加到vector<node> link中；
修改link中链表信息，输出结果
注意点：在使用vector的时候，如果事先申明了vector的大小，如果再使用push_back()在压在申请内存的后面的，而不是从头开始压入的； 
　　　　此外，pat的链表类题中，经常存在不在链表中的节点， 所以在最后的循环输出的时候，不能以节点个数作为循环结束的判断，应该以link的大小作为循环次数的判断

#include<iostream>
#include<vector>
using namespace std;
struct Node{
  int addr, val, next;
};
int main(){
  int root, n, k, i, j;
  scanf("%d %d %d", &root, &n, &k);
  vector<Node> linkList(100000);
  for(i=0; i<n; i++){
    int addr, val, next;
    scanf("%d %d %d", &addr, &val, &next);
    linkList[addr].addr=addr; linkList[addr].val = val; linkList[addr].next = next;
  }
  int cnt=(n+k-1)/k, nodeCnt=0;
  vector<vector<int> > idx(cnt);
  while(root!=-1){//找出链表中所有节点的位置
    if(idx[nodeCnt].size()>=k) nodeCnt++;
    idx[nodeCnt].push_back(root);
    root = linkList[root].next;
  }
  for(i=0; i<cnt-1; i++){//先输出前面完整的分组
    int len=idx[i].size();
    for(j=len-1; j>0; j--){
      int index=idx[i][j], preIndex=idx[i][j-1];
      if(j!=0) printf("%05d %d %05d
", index, linkList[index].val, preIndex);
    }
    if(idx[i+1].size()==k) printf("%05d %d %05d
", idx[i][0], linkList[idx[i][0]].val, idx[i+1][k-1]);
    else printf("%05d %d %05d
", idx[i][0], linkList[idx[i][0]].val, idx[i+1][0]);
  }
  if(idx[cnt-1].size()==k){//处理最后一个分组
    for(j=k-1; j>0; j--) printf("%05d %d %05d
", idx[cnt-1][j], linkList[idx[cnt-1][j]].val, idx[cnt-1][j-1]);
    printf("%05d %d %d
", idx[cnt-1][0], linkList[idx[cnt-1][0]].val, -1);
  }else{
    for(i=0; i<idx[cnt-1].size()-1; i++) 
      printf("%05d %d %05d
", idx[cnt-1][i], linkList[idx[cnt-1][i]].val, idx[cnt-1][i+1]);
    printf("%05d %d %d
", idx[cnt-1][i], linkList[idx[cnt-1][i]].val, -1);  
  }
  return 0;
}