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 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

算法如下：

#include<stdio.h>  #include<stdlib.h>  typedef struct Node  {      int curadd;  //当前地址    int data;      int nextadd;  //下个节点地址}Node;  int main()  {      int n,k,firstadd,i,num,j;  //n为输入节点个数，k为逆序位元    Node *a,temp;      int c; // 保存逆序的组数    while(scanf("%d%d%d",&firstadd,&n,&k)!=EOF)      {   //创建一个节点数组        a=(Node *)malloc(n*sizeof(Node));  //对数组元素进行赋值        for(i=0;i<n;i++)              scanf("%d%d%d",&a[i].curadd,&a[i].data,&a[i].nextadd);          num=0;  //将输入节点按地址顺序化        while(firstadd!=-1)          {              j=num; //找到第一个节点            while(a[j].curadd!=firstadd)                  j++;              firstadd=a[j].nextadd;//交换num与j所在的节点位置            temp=a[num];              a[num]=a[j];              a[j]=temp;              num++;          }          if(k>num)  //逆序位元大于节点数直接输出        {              for(i=0;i<num-1;i++)                  printf("%05d %d %05d\n",a[i].curadd,a[i].data,a[i+1].curadd);              printf("%05d %d -1\n",a[i].curadd,a[i].data);          }          else if(k==num)  //逆序位元等于节点数全部逆序输出        {              for(i=num-1;i>0;i--)                  printf("%05d %d %05d\n",a[i].curadd,a[i].data,a[i-1].curadd);              printf("%05d %d -1\n",a[0].curadd,a[0].data);          }          else  //进行前k个逆序的逆序        {              c=num/k; //调整顺序//逆序组的循环            for(i=0;i<c;i++)              {  //组内循环交换k*i+j(组内第一个)和k*i+k-j-1(组内最后一个)对应的节点                for(j=0;j<k/2;j++)                  {                      temp=a[k*i+j];                      a[k*i+j]=a[k*i+k-j-1];                      a[k*i+k-j-1]=temp;                  }              }              for(i=0;i<num-1;i++)                  printf("%05d %d %05d\n",a[i].curadd,a[i].data,a[i+1].curadd);              printf("%05d %d -1\n",a[i].curadd,a[i].data);          }      }      return 0;  }  

下面给出C++的代码实现，此算法优于上面的算法：

#include <vector>#include <iostream>#include <string.h>#include <algorithm> using namespace std; struct Node{    int addr;    int a;    int next;}; int main(){  //  输入        // 读入第一行    int root;    int n;    int k;    cin >> root >> n >> k;      // 读入其余行    vector<Node> inLst;<span style="white-space:pre"></span>int posOf[100005];//下标代表地址测试地址为5位，故100000则够用，数组元素为各个节点在inLst中的位置    int cnt = 0;    memset(posOf, -1, sizeof(posOf)); //方便以后的处理，将数组所有元素置为-1     for (int i = 0; i < n; i++) {        Node node;        cin >> node.addr >> node.a >> node.next;             inLst.push_back(node);        posOf[node.addr] = cnt++;      }//  遍历    vector<Node> linkLst; //用于保存按顺序排好的链表；     int p = root;    for (int i = 0; i < n; ++i) {        Node node = inLst[posOf[p]];        linkLst.push_back(node);                 p = node.next;                 if (p == -1) {            break;        }    }     //  逆置     int time = (int)linkLst.size() / k;//time为需要逆置的组数     for (int i = 0; i < time; i++) {        reverse(linkLst.begin() + i * k, linkLst.begin() + (i + 1) * k);//分别代入逆置头和尾的迭代器     }//  输出    int size = (int)linkLst.size();    for (int i = 0; i < linkLst.size(); i++){        if (i == size - 1) {            printf("%05d %d -1\n",linkLst[i].addr,linkLst[i].a);        }else{            printf("%05d %d %05d\n",linkLst[i].addr,linkLst[i].a,linkLst[i+1].addr);        }    }    return 0;}