PAT乙级真题及训练题 1025. 反转链表 (25)

PAT乙级真题及训练题 1025. 反转链表 (25)

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感觉几个世纪没打代码了，真是坏习惯，调了两小时把反转链表调出来了，心情舒畅。

这道题的步骤

• 数据输入，数组纪录下一结点及储存值
• 创建链表并储存上下结点“位置”
• 按照规律遍历一遍链表并反转链表
• 输出整条反转完的链表

#include <iostream>
#include <cstdio>
using namespace std;
int nextnum[100010],value[100010];
struct Node{
    int value,pos,nnext;
    Node *pnext,*plast,*final;
};
int main() {
    int start,n,k,pos,next,val,count=1;
    
    
    
    /*-------------数据输入，数组纪录下一结点及储存值-------------*/
    scanf("%d%d%d",&start,&n,&k);
    for (int i=0; i<n; i++) {
        scanf("%d%d%d",&pos,&val,&next);
        nextnum[pos]=next;
        value[pos]=val;
    }
    /*------------------------------------------------------*/
    
    
    
    
    
    /*-------------创建链表并储存上下结点“位置”-----------------*/
    pos=start;
    Node *p1,*p2,*head;
    p1=p2=new Node;
    p1->value=value[pos];
    p1->pos=pos;
    head=p1;
    pos=nextnum[pos];
    while (pos!=-1) {
        p1=new Node;
        p2->pnext=p1;
        p1->plast=p2;
        p1->value=value[pos];
        p1->pos=pos;
        p2=p1;
        pos=nextnum[pos];
        count++;
    }
    p1->pnext=NULL;

    if (count==1) {
        printf("%05d %d -1\n",head->pos,head->value);
        return 0;
    }
    /*------------------------------------------------------*/
    
    
    
    
    
    /*-------------反转链表----------------------------------*/
    int cur=0;
    Node *temp,*temp1=NULL;
    temp=head;
    p1=head;
    while (p1) {
        if (cur==k-1) {
            head=p1;
        }//find the head
        
        if (cur%k==0 && cur<count/k*k) {
            p1->nnext=-1;
            p1->final=NULL;
            if (cur%(2*k)==0) temp=p1;
            else temp1=p1;
        }else {
            if (cur<count/k*k) {
                p1->final=p1->plast;
                p1->nnext=p1->final->pos;
            }else {
                p1->final=p1->pnext;
                if (cur<count-1) p1->nnext=p1->final->pos;
                else p1->nnext=-1;
            }
        }
        
        if (cur>k-1 && (cur+1)%k==0) {
            if ((cur+1)%(2*k)==0) {
                temp->final=p1;
                temp->nnext=p1->pos;
            }else {
                temp1->final=p1;
                temp1->nnext=p1->pos;
            }
            
        }
        if (cur==count/k*k) {
            if (cur%(2*k)!=0) {
                temp->final=p1;
                temp->nnext=p1->pos;
            }else {
                temp1->final=p1;
                temp1->nnext=p1->pos;
            }
            
        }
        
        cur++;
        p1=p1->pnext;
        if (cur==count) {
            break;
        }
    }
    /*------------------------------------------------------*/
    
    
    
    
    
    /*-------------输出整条反转完的链表------------------------*/
    p1=head;
    while (p1) {
        if (p1->nnext==-1) printf("%05d %d %d\n",p1->pos,p1->value,p1->nnext);
        else printf("%05d %d %05d\n",p1->pos,p1->value,p1->nnext);
        p1=p1->final;
    }
    
    
    return 0;
}

另贴出一两个月前用指针结合数组的做法做的，作为对比

#include <iostream>  
#include <cstdio>  
using namespace std;  
struct Node{  
    Node * pointer;  
    int nnext;  
    int position;  
    int nvalue;  
};  
Node line[100010],input[100010];  
int main(){  
    int first,n,mol,  
        num,value,next,  
        i;  
    cin >> first >> n >> mol;  
    if (n==1) {  
        cin >> num >> value >> next;  
        printf("%05d %d -1\n",num,value);  
    }else{  
        for (i=0; i<n; i++) {  
            cin >> num >> value >> next;  
            if (next == -1) next=100002;  
            input[num].nvalue = value;  
            input[num].nnext = next;  
            input[num].position = num;  
            if (num == first) line[0] = input[num];  
        }  
        for (i=1; i<n; i++) {  
            line[i]=input[line[i-1].nnext];  
            if (line[i].nnext == 100002) break;  
        }  
        n=i+1;  
        int nreverse=(n/mol)*mol;  
        for (i=0; i<nreverse; i++) {  
            if (i%mol!=0) line[i].pointer = &line[i-1];  
            else if (i==nreverse-mol) line[i].pointer = &line[nreverse];  
            else line[i].pointer = &line[i+2*mol-1];  
            line[i].nnext = line[i].pointer->position;  
        }  
        for (i=nreverse; i<n-1; i++) line[i].pointer = &line[i+1];  
        if (n%mol == 0) {  
            line[n-mol].pointer = NULL;  
            line[n-mol].nnext = 100002;  
        }  
        else line[n-1].pointer = NULL;  
        Node * p= &line[mol-1];  
        if (n<mol) p = &line[0];  
        while (p != NULL) {  
            if (p->nnext != 100002) printf("%05d %d %05d\n",p->position,p->nvalue,p->nnext);  
            else printf("%05d %d -1\n",p->position,p->nvalue);  
            p = p->pointer;  
        }  
    }  
    return 0;  
}