02-线性结构2 Reversing Linked List

Posted 2020-06-21 kuotian

由于最近学的是线性结构，且因数组需开辟的空间太大。因此这里用的是纯链表实现的这个链表翻转。

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 NN 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 Nextis 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

 

  1 #include <stdio.h>
  2 #include <stdlib.h>
  3 
  4 typedef struct Node
  5 {
  6     int address;
  7     int data;
  8     int nextAddress;
  9     struct Node *next;
 10 }Node;
 11 typedef struct Node *LinkList;
 12 
 13 int main()
 14 {
 15     //排序前 
 16     LinkList L1, p1, q1;            
 17     L1 = (LinkList)malloc(sizeof(Node));    //创建头指针 
 18     L1->next = NULL;
 19     int firstAddress;
 20     int N, K;//N为总结点数 K为需翻转的数 
 21     scanf("%d %d %d", &firstAddress, &N, &K);
 22     p1 = L1;
 23     for(int i = 0; i < N; i++) {
 24         q1 =  (LinkList)malloc(sizeof(Node));
 25         scanf("%d %d %d",&q1->address, &q1->data, &q1->nextAddress);
 26         p1->next = q1;
 27         p1 = q1;
 28     }
 29     p1->next = NULL;
 30     
 31 //    //测试没问题 
 32 //    printf("测试1 ：\n");
 33 //    p1 = L1->next;
 34 //    while(p1){
 35 //        printf("%05d %d %d\n", p1->address, p1->data, p1->nextAddress);
 36 //        p1 = p1->next;
 37 //    }
 38     
 39     //排序后 
 40     LinkList L2, p2;
 41     L2 = (LinkList)malloc(sizeof(Node));    //创建头指针 
 42     L2->next = NULL;
 43     int count = 0;
 44     int findAddress = firstAddress;
 45     p2 = L2;
 46     while(findAddress != -1) {            //while(count < N) {有多余结点不在链表上没通过 
 47     
 48         q1 = L1;
 49         while(q1->next) {
 50             if(q1->next->address == findAddress) {
 51                 p2->next = q1->next;
 52                 q1->next = q1->next->next;
 53                 p2 = p2->next;
 54                 count++;
 55 //                printf("count = %d\n",count);
 56                 findAddress = p2->nextAddress;
 57 //                printf("findAddress = %d\n",findAddress);
 58             }else {
 59                 q1 = q1->next;
 60             }
 61         }
 62     }
 63     p2->next = NULL;
 64     
 65 //    //测试没问题 
 66 //    printf("测试2 ：\n");
 67 //    p2 = L2->next;
 68 //    while(p2){
 69 //        printf("%05d %d %05d\n", p2->address, p2->data, p2->nextAddress);
 70 //        p2 = p2->next;
 71 //    }
 72     //Reversing
 73     LinkList L3, p3, q3, tail;
 74     L3 = (LinkList)malloc(sizeof(Node));    //创建头指针 
 75     L3->next = NULL;
 76     //将L2以头插法插入L3
 77     int n = count;                //防止有多余结点影响 n=N 会影响
 78     int k = K;
 79     p3 = L3;
 80     p2 = L2;
 81     while(n >= k) {
 82         n -= k;
 83         for(int i = 0; i < k; i++) {
 84             p3->next = p2->next;
 85             p2->next = p2->next->next;
 86             if(i == 0)
 87                 tail = p3->next;
 88             else 
 89                 p3->next->next = q3;
 90             q3 = p3->next;
 91         }
 92         p3 = tail;
 93     }
 94     p3->next = L2->next;
 95     
 96     p3 = L3->next;
 97     while(p3->next) {
 98         printf("%05d %d %05d\n",p3->address, p3->data, p3->next->address);//不到五位数用0补全 
 99         p3 = p3->next;
100     }
101     printf("%05d %d -1\n",p3->address, p3->data);
102     return 0;
103 }