02-线性结构3 Reversing Linked List (25 分)

如果单纯从做题角度来讲，上一个做法当然没问题，可是目的是为了练习静态链表，so还是不要投机取巧了，静态链表解法如下：

#include <stdio.h>
struct data
{int n;int next;} ss[100000];int find(int first);int reverse(int first,int sum,int k);void print(int first);int main(){int i,j;int first,sum,k;scanf("%d%d%d",&first,&sum,&k);for(i=0;i<sum;i++){scanf("%d",&j);scanf("%d%d",&ss[j].n,&ss[j].next);}sum=find(first);if(k>1){first=reverse(first,sum,k); }print(first);return 0;}int find(int first){int i=1;while(ss[first].next!=-1){first=ss[first].next;i++;}return i;}int reverse(int first,int sum,int k)//只解释这个主要函数。 {int i;int head;int p1,p2,p3;//p1,p2,p3轮替使指针移位，并且交换指针。 int firstend=first;//记录上一次的末尾，以便跟这一次的头相连。 for(i=0;i<sum/k;i++){int key=k;key--;p2=first;p3=ss[first].next;while(key--)//循环交换。 {p1=p2;p2=p3;p3=ss[p2].next;ss[p2].next=p1;}if(i==0)//第一次时无需连接上一次的未和这一次的头，且要返回新的头指针 {head=p2;if(p3!=-1){first=p3;}else{ss[first].next=-1;}}else{ss[firstend].next=p2;firstend=first;if(p3!=-1){first=p3;}else{ss[first].next=-1;}}}if(p3!=-1)//跳出for循环而不空，直接相连即可。 {ss[firstend].next=p3;}return head;}void print(int first){while(ss[first].next!=-1){printf("%05d %d %05d\n",first,ss[first].n,ss[first].next);first=ss[first].next;}printf("%05d %d %d\n",first,ss[first].n,ss[first].next);}

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02-线性结构3 Reversing Linked List (25 分)

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