题目地址

https://pta.patest.cn/pta/test/16/exam/4/question/664

 

5-2 Reversing Linked List   (25分)

Given a constant KK and a singly linked list LL, you are supposed to reverse the links of every KK elements on LL. For example, given LL being 1→2→3→4→5→6, if K = 3K=3, then you must output 3→2→1→6→5→4; if K = 4K=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 NN (\le 10^5≤10​5​​) which is the total number of nodes, and a positive KK (\le N≤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 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

/*评测结果时间	结果	得分	题目	编译器	用时（ms）	内存（MB）	用户2017-07-08 15:54	答案正确	25	5-2	gcc	131	3	测试点结果测试点	结果	得分/满分	用时（ms）	内存（MB）测试点1	答案正确	12/12	2	1测试点2	答案正确	3/3	14	1测试点3	答案正确	2/2	1	1测试点4	答案正确	2/2	2	1测试点5	答案正确	2/2	2	1测试点6	答案正确	3/3	131	3测试点7	答案正确	1/1	2	1*/#include
       
        #define MAXLEN 100002struct node {	int data;	int next;};int k,head;struct node workArray [MAXLEN];int Input(struct node  array[]){	int i,inputHead,inputLength;	int index,data,next;		scanf("%d %d %d",&inputHead,&inputLength,&k);	for (i=0;i
        
         ptr2改为ptr2->ptr1。因为ptr2中的next原有内容会丢失，故用ptr3保存ptr2的下一个节点	执行完一次后，k个节点区间内，头尾互换。	故lastend保存前一区块的末端，是上一区间的头节点	nexthead即下一区块的头结点，同样也是该区块翻转完后的末端。于是提前用lastend=nexthead保存。 	*/	int cnt;	if(k==1)		return;	cnt=count(*head,array);	int i,ptr1,ptr2,ptr3,firstflag=0,nexthead=*head,lastend=-2;//ptr1指当前指针，ptr2指下一个要指向ptr1的，ptr3指向还未做反转的下一个。	while(cnt>=k){//		printf("-------head=%d,nexthead=%d,cnt=%d\n",*head,nexthead,cnt);//for_test		ptr1=nexthead;		ptr2=array[ptr1].next;		for(i=1;i