PAT甲级-链表处理-1074 Reversing Linked List解题思路

1074 Reversing Linked List (25 分)

思路

链表好坑，每k个要翻转一次，样例明显没有啊
而且还有游走的指针，不在链表内，需要统计合法链表内的指针个数

细节，就是开链表大小的结构数组去作为链表地址去存储，然后按照顺序给予他们一个rank，这里的rank可以通过结构体rank给予 或 使用rank数组进行标定

代码一（纯链表，加指针翻转）

#include<bits/stdc++.h>
using namespace std;
const int maxn = 100010;
int start ,n,k;
struct Node{
    
    
    int address,data,next;
    int order;
}nodes[maxn];

bool cmp(Node a, Node b)
{
    
    
    return a.order<b.order;
}

int main(){
    
    
    cin>>start>>n>>k;
    for(int i = 0;i<maxn;i++){
    
    
        nodes[i].order = maxn;
    }

    int add ;
    for(int i = 0;i<n;i++)
    {
    
    
        cin>>add;
        cin>>nodes[add].data>>nodes[add].next;
        nodes[add].address = add;
    }

    int count = 0;
    int p =start;
    while( p != -1)
    {
    
    
        nodes[p].order =count++;
        p=nodes[p].next;
    }
    sort(nodes,nodes+maxn,cmp);

    if(k>count) k=count;
    int ttt = k;
    if(k>1)
    {
    
    
        while(ttt<=count)
        {
    
    
            int begin = nodes[ttt-1].next;
            for(int i=ttt-1;i>ttt-k;i--)
                nodes[i].next = nodes[i-1].address;
            nodes[ttt-k].next = begin;
            
            if(ttt+k <= count)
                nodes[ttt-k].next = nodes[ttt+k-1].address;
            for(int i=ttt-1;i>=ttt-k;i--)
            {
    
       if(nodes[i].next == -1)
                    printf("%05d %d %d\n",nodes[i].address,nodes[i].data,nodes[i].next);
                else 
                printf("%05d %d %05d\n",nodes[i].address,nodes[i].data,nodes[i].next);
            }

            ttt+=k;
        }
        ttt-=k;
    }
    else{
    
    ttt = 0;}
    
    for(int i=ttt;i<count;i++)
    {
    
       if(nodes[i].next == -1)
            printf("%05d %d %d\n",nodes[i].address,nodes[i].data,nodes[i].next);
        else 
            printf("%05d %d %05d\n",nodes[i].address,nodes[i].data,nodes[i].next);
    }

}

代码二（借助数组）

#include<bits/stdc++.h>
using namespace std;
const int maxn = 100010;
int start ,n,k;


int next_[maxn],data[maxn],rank_[maxn]={
    
    0};
int main()
{
    
    
    cin>>start>>n>>k;
    int add;
    for(int i=0;i<n;i++)
    {
    
    
        cin>>add;
        cin>>data[add]>>next_[add];
    }
    int count = 0,p=start;
    while(p!=-1)
    {
    
    
        rank_[count++] =  p;
        p = next_[p];
    }
    for(int i = 0;i<count-count%k;i+=k)
        reverse(rank_+i,rank_+k+i);
    for(int i=0;i<count-1;i++)
        printf("%05d %d %05d\n",rank_[i],data[rank_[i]],rank_[i+1]);
    printf("%05d %d -1\n",rank_[count-1],data[rank_[count-1]]);

}