02-线性结构3 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

最初版本代码，建立原始链表采用N次循环导致超时，没有采用格式化输出，用了很傻的方法。。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <malloc.h>
using namespace std;
struct point
{int prev;int data;int next;
};
struct node{int addr;int data;struct node *next;
};struct node *Rverse(struct node *p,int k,int n){if(k==1) return p;else{struct node *temp,*b,*c,*d,*first;first=p;temp=p->next;b=temp->next;if(b->next){while(temp->next){int cnt=0;while(cnt<k && b->next){c=b->next;b->next=temp;temp=b;b=c;cnt++;}if(cnt+2<k){while(cnt>0){c=temp->next;temp->next=b;b=temp;temp=c;cnt--;}temp->next=b;first->next=temp;break;}else if(cnt==k){d=first->next;first->next=temp->next;first=d;first->next=temp;}else if(cnt+2==k){b->next=temp;temp=first->next;temp->next=0;first->next=b;break;}}}else{first->next=b;b->next=temp;}return p;}
}
void printAddr(int n){if(n>9999) cout<<n;else if(n>999) cout<<0<<n;else if(n>99) cout<<0<<0<<n;else if(n>9) cout<<0<<0<<0<<n;else cout<<0<<0<<0<<0<<n;
}int main()
{int firstAddr,n,k,i,exit=0;cin>>firstAddr>>n>>k;struct point oriList[n];for(i=0;i<n;i++){cin>>oriList[i].prev>>oriList[i].data>>oriList[i].next;}struct node *head,*p;head=(struct node *)malloc(sizeof(struct node));p=(struct node *)malloc(sizeof(struct node));head->next=p;p->addr=firstAddr;while(1){struct node *p1;p1=(struct node *)malloc(sizeof(struct node));for(int i=0;i<n;i++){if(oriList[i].prev==p->addr){p->data=oriList[i].data;if(oriList[i].next==-1){exit=1;break;}p1->addr=oriList[i].next;p->next=p1;p=p1;break;}}if(exit) break;}p->next=0;p=head;p=Rverse(p,k,n);while(1){p=p->next;printAddr(p->addr);cout<<' ';cout<<p->data<<' ';if(p->next){printAddr(p->next->addr);cout<<endl;}else{cout<<-1<<endl;break;}}return 0;
}

建立单链表的时候使用了二分查找和插值查找，还是超时。。代码如下：

#include <iostream>
#include <cstring>
#include <cstdio>
#include<algorithm>
#include<fstream>
#include <malloc.h>
using namespace std;
struct point
{int prev;int data;int next;
};
bool cmp(const point &a,const point &b)
{if(a.prev!=a.prev) return a.prev<b.prev;else return a.prev<b.prev;
}
typedef struct node{int addr;int data;struct node *next;
}node;node *Rverse(node *p,int k,int n){if(k==1) return p;else{node *temp,*b,*c,*d,*first;first=p;temp=p->next;b=temp->next;if(b->next){while(temp->next){int cnt=0;while(cnt<k && b->next){c=b->next;b->next=temp;temp=b;b=c;cnt++;}if(cnt+2<k){while(cnt>0){c=temp->next;temp->next=b;b=temp;temp=c;cnt--;}temp->next=b;first->next=temp;break;}else if(cnt==k){d=first->next;first->next=temp->next;first=d;first->next=temp;}else if(cnt+2==k){b->next=temp;temp=first->next;temp->next=0;first->next=b;break;}}}else{first->next=b;b->next=temp;}return p;}
}
void printAddr(int n){if(n>9999) cout<<n;else if(n>999) cout<<0<<n;else if(n>99) cout<<0<<0<<n;else if(n>9) cout<<0<<0<<0<<n;else cout<<0<<0<<0<<0<<n;
}int main()
{int firstAddr,n,k,i,exit=0;cin>>firstAddr>>n>>k;struct point oriList[n];for(i=0;i<n;i++){cin>>oriList[i].prev>>oriList[i].data>>oriList[i].next;}sort(oriList, oriList+n,cmp);struct node *head,*p;head=(node *)malloc(sizeof(node));p=(node *)malloc(sizeof(node));head->next=p;p->addr=firstAddr;while(1){int minb=0,maxb=n-1;struct node *p1;p1=(node *)malloc(sizeof(node));while(minb<=maxb){int mid;if(maxb-minb>50){mid=(p->addr-oriList[minb].prev)/(oriList[maxb].prev-oriList[minb].prev)*(maxb-minb);}else{mid=(minb+maxb)/2;}if(p->addr>oriList[mid].prev){minb=mid+1;}else if(p->addr<oriList[mid].prev){maxb=mid-1;}else{p->data=oriList[mid].data;if(oriList[mid].next==-1){exit=1;break;}p1->addr=oriList[mid].next;p->next=p1;p=p1;break;}}if(exit) break;}p->next=0;p=head;p=Rverse(p,k,n);while(1){p=p->next;printAddr(p->addr);cout<<' ';cout<<p->data<<' ';if(p->next){printAddr(p->next->addr);cout<<endl;}else{cout<<-1<<endl;break;}}return 0;
}

借鉴大神的代码如下：

#include<stdio.h>
#define MAX_SIZE 100004typedef struct tagLNode{int addr;      //节点位置Addressint data;      //Data值int nextAddr;  //下个节点位置struct tagLNode *next;  //指向下个节点的指针
} LNode;
/*LNode *listReverse(LNode *head, int k);  反转单链表函数参数1：单链表的头节点，参数2：反转子链表的长度，返回值：反转后的链表的第一个节点(不是头结点)*/
LNode *listReverse(LNode *head, int k);
//输出单链表 参数为单链表的头结点
void printList(LNode *a);int main()
{int firstAddr;int n = 0;            //节点数 Nint k = 0;            //反转子链表的长度Kint num = 0;          //链表建好之后的链表节点数int data[MAX_SIZE];   //存data值 节点位置作为索引值int next[MAX_SIZE];   //存next值 节点位置为索引int tmp;              //临时变量，输入的时候用scanf("%d %d %d", &firstAddr, &n, &k);    LNode a[n+1];                //能存n+1个几点的数组。a[0].nextAddr = firstAddr;   //a[0] 作为头节点//读输入的节点int i = 1;    for (; i < n+1; i++){scanf("%d", &tmp);scanf("%d %d", &data[tmp], &next[tmp]);        }//构建单链表i = 1;while (1){if (a[i-1].nextAddr == -1){a[i-1].next = NULL;num = i-1;break;            }a[i].addr = a[i-1].nextAddr;a[i].data = data[a[i].addr]; a[i].nextAddr = next[a[i].addr];a[i-1].next = a+i;i++;}LNode *p = a;                    //p指向链表头结点LNode *rp = NULL;                //反转链表函数的返回值if (k <= num ){for (i = 0; i < (num/k); i++){rp = listReverse(p, k);  //p->next = rp;            // 第一次执行，a[0]->next 指向第一段子链表反转的第一个节点p->nextAddr = rp->addr;  // 更改Next值，指向逆转后它的下一个节点的位置int j = 0;//使p指向下一段需要反转的子链表的头结点（第一个节点的前一个节点）while (j < k){p = p->next;j++;}}}printList(a);
}LNode *listReverse(LNode *head, int k)
{int count = 1;LNode *new = head->next;LNode *old = new->next;LNode *tmp = NULL;while (count < k){tmp = old->next;old->next = new;old->nextAddr = new->addr;new = old;   //new向后走一个节点old = tmp;   //tmp向后走一个节点count++;}//使反转后的最后一个节点指向下一段子链表的第一个节点head->next->next = old;  if (old != NULL){//修改Next值，使它指向下一个节点的位置head->next->nextAddr = old->addr; }else{//如果old为NULL,即没有下一个子链表，那么反转后的最后一个节点即是真个链表的最后一个节点head->next->nextAddr = -1;       }return new;
}void printList(LNode *a)
{LNode *p = a;while (p->next != NULL){p = p->next;        if (p->nextAddr != -1 ){//格式输出，%.5意味着如果一个整数不足5位，输出时前面补0 如：22，输出：00022printf("%.5d %d %.5d\n", p->addr, p->data, p->nextAddr);}else{//-1不需要以%.5格式输出printf("%.5d %d %d\n", p->addr, p->data, p->nextAddr);}    }}