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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 }
02-线性结构2 Reversing Linked List
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原文地址：http://www.cnblogs.com/kuotian/p/5269434.html
