02 - 线性结构 2 Reversing Linked List (25 分)

 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
 

#include 
 
 
define MAX 100000
 
 
typedef struct{
 int data;
 int next;
 }Node;
 //计算并返回数组中在链表中的结点
 int CountNodes(Node*list, int pList) {
 int cnt = 1;
 while((pList = list[pList].next) != -1) cnt++;
 return cnt;
 }
 //list指向数组 plist为第一个元素的地址 n为总数 k为反转单位
 int ReverseK(Node*list, int pList, int n, int k) {
 //前一节点，当前节点，下一节点的地址
 int prevNode, currNode, nextNode;
 prevNode = -1;
 currNode = pList;
 nextNode = list[currNode].next;
 int lastHead, head = -1;
 int i ,j;
 for( i = 0; i < n / k; i++){
 lastHead = head;//记录前一段的（未逆转的）头结点，以便连接到当前段的(未逆转的）尾节点
 head = currNode;//记录当前段的头结点
 for(j = 0; j < k; j++) { //k个结点逆转，后一节点指向前一节点
 list[currNode].next = prevNode;
 prevNode = currNode;
 currNode = nextNode;
 nextNode = list[nextNode].next;
 }
 if(i == 0)//第一段逆转后的头结点当作表头返回
 pList = prevNode;
 else
 list[lastHead].next = prevNode;//连接逆转后的前后两段
 }
 list[head].next = currNode; //不用逆转的剩余部分加上
 return pList;
 }
 void printList(Node* list, int p) {
 while(list[p].next != -1){
 printf("%05d %d %05d\n", p, list[p].data, list[p].next);
 p = list[p].next;
 }
 printf("%05d %d %d\n", p, list[p].data, list[p].next);
 }
 int main() {
 int pList, n, k;
 scanf("%d%d%d", &pList, &n, &k);
 Node list[MAX];
 int addr, data, next,i;
 for(i = 0; i < n; i++){
 scanf("%d%d%d", &addr, &data, &next);
 list[addr].data = data;
 list[addr].next = next;
 }
 int num = CountNodes(list, pList);
 int pNewList = ReverseK(list, pList, num, k);
 printList(list, pNewList);
 return 0;
 }