数组模拟链表运作！！L2-002. 链表去重

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给定一个带整数键值的单链表L，本题要求你编写程序，删除那些键值的绝对值有重复的结点。即对任意键值K，只有键值或其绝对值等于K的第一个结点可以被保留。同时，所有被删除的结点必须被保存在另外一个链表中。例如：另L为21→-15→-15→-7→15，则你必须输出去重后的链表21→-15→-7、以及被删除的链表-15→15。

输入格式：

输入第一行包含链表第一个结点的地址、以及结点个数N（<= 10⁵ 的正整数）。结点地址是一个非负的5位整数，NULL指针用-1表示。

随后N行，每行按下列格式给出一个结点的信息：

Address Key Next

其中Address是结点的地址，Key是绝对值不超过10⁴的整数，Next是下一个结点的地址。

输出格式：

首先输出去重后的链表，然后输出被删除结点组成的链表。每个结点占一行，按输入的格式输出。

输入样例：

00100 599999 -7 8765423854 -15 0000087654 15 -100000 -15 9999900100 21 23854

输出样例：

00100 21 2385423854 -15 9999999999 -7 -100000 -15 8765487654 15 -1#include<stdio.h>#include<string.h>#include<algorithm>using namespace std; int vis[1001100];    int mapp[1001100],mappp[1001100];    int keep1[1001100],keep2[1001100];int main(){    int a,b,c;    int vv1=0,vv2=0;    memset(vis,0,sizeof(vis));    int m,n;    scanf("%d%d",&m,&n);    for(int i=0;i<n;i++)    {        scanf("%d%d%d",&a,&b,&c);        mapp[a]=b;        mappp[a]=c;    }    int pp;    pp=m;   while(mappp[pp]!=-1)   {       int flag=0;       if(mapp[pp]<0)       {           flag=1;           mapp[pp]=-mapp[pp];       }       if(vis[mapp[pp]]==0)       {           if(vv2>0)            keep2[vv2-1]=mappp[pp];           if(flag==1)           {           keep1[vv1++]=pp;           keep1[vv1++]=-mapp[pp];           keep1[vv1++]=mappp[pp];           vis[mapp[pp]]=1;           pp=mappp[pp];           }          else          {           keep1[vv1++]=pp;           keep1[vv1++]=mapp[pp];           keep1[vv1++]=mappp[pp];           vis[mapp[pp]]=1;           pp=mappp[pp];          }       }       else       {           keep1[vv1-1]=mappp[pp];           if(flag==1)           {           keep2[vv2++]=pp;           keep2[vv2++]=-mapp[pp];           keep2[vv2++]=mappp[pp];           pp=mappp[pp];           }           else           {           keep2[vv2++]=pp;           keep2[vv2++]=mapp[pp];           keep2[vv2++]=mappp[pp];           pp=mappp[pp];           }       }   }   int flag1=0;   if(mapp[pp]<0)   {       flag1=1;       mapp[pp]=-mapp[pp];   }   if(vis[mapp[pp]]==1)   {       if(flag1==0)       {           keep2[vv2++]=pp;           keep2[vv2++]=mapp[pp];           keep2[vv2++]=mappp[pp];           keep1[vv1-1]=-1;       }       else       {           keep2[vv2++]=pp;           keep2[vv2++]=-mapp[pp];           keep2[vv2++]=mappp[pp];           keep1[vv1-1]=-1;       }   }   else   {       if(!flag1)       {           keep1[vv1++]=pp;           keep1[vv1++]=mapp[pp];           keep1[vv1++]=mappp[pp];           keep2[vv2-1]=-1;       }       else       {           keep1[vv1++]=pp;           keep1[vv1++]=-mapp[pp];           keep1[vv1++]=mappp[pp];           keep2[vv2-1]=-1;       }   }   for(int i=1;i<=vv1;i++)   {       if(keep1[i-1]==-1)        printf("-1");       else if(i%3==1||i%3==0)        printf("%05d",keep1[i-1]);         else if(i%3==2)        printf("%d",keep1[i-1]);        if(i%3)            printf(" ");        else            printf("\n");   }   for(int i=1;i<=vv2;i++)   {       if(keep2[i-1]==-1)        printf("-1");       else if(i%3==1||i%3==0)        printf("%05d",keep2[i-1]);       else if(i%3==2)        printf("%d",keep2[i-1]);        if(i%3)            printf(" ");        else            printf("\n");   }    return 0;}

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