02-线性结构3 Reversing Linked List

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 100004

typedef struct tagLNode{
    int addr;      //节点位置Address
    int 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 N
    int k = 0;            //反转子链表的长度K
    int 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：00022
            printf("%.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);
        }    
    }

}