1097 Deduplication on a Linked List——PAT甲级

Given a singly linked list L with integer keys, you are supposed to remove the nodes with duplicated absolute values of the keys. That is, for each value K, only the first node of which the value or absolute value of its key equals K will be kept. At the mean time, all the removed nodes must be kept in a separate list. For example, given L being 21→-15→-15→-7→15, you must output 21→-15→-7, and the removed list -15→15.

Input Specification:

Each input file contains one test case. For each case, the first line contains the address of the first node, and a positive N (≤105) which is the total number of nodes. 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:

复制代码
Address Key Next

where Address is the position of the node, Key is an integer of which absolute value is no more than 104, and Next is the position of the next node.

Output Specification:

For each case, output the resulting linked list first, then the removed list. Each node occupies a line, and is printed in the same format as in the input.

Sample Input:

复制代码
00100 5
99999 -7 87654
23854 -15 00000
87654 15 -1
00000 -15 99999
00100 21 23854

Sample Output:

复制代码
00100 21 23854
23854 -15 99999
99999 -7 -1
00000 -15 87654
87654 15 -1

solution:

cpp 复制代码
#include<bits/stdc++.h>
using namespace std;
const int maxn=1e5+5;
int Next[maxn]={0};
int Data[maxn]={0};
int main()
{
    int head,n;
    cin>>head>>n;
    for(int i=0;i<n;i++)
    {
        int t;cin>>t;
        cin>>Data[t]>>Next[t];
    }
    vector<pair<int,int> >l1,ans,del;
    int cnt=0;
    while(head!=-1)
    {
        cnt++;
        l1.push_back({head,Data[head]});
        head=Next[head];
    }
    set<int>s;
    for(int i=0;i<cnt;i++)
    {
        if(s.count(abs(l1[i].second)))
        {
            del.push_back({l1[i].first,l1[i].second});
            continue;
        }
        ans.push_back({l1[i].first,l1[i].second});
        s.insert(abs(l1[i].second));
    }
    for(int i=0;i<ans.size()-1;i++)
    {
        printf("%05d %d %05d\n",ans[i].first,ans[i].second,ans[i+1].first);
    }
    printf("%05d %d -1\n",ans[ans.size()-1].first,ans[ans.size()-1].second);
    if(del.size())
    {
        for(int i=0;i<del.size()-1;i++)
        {
            printf("%05d %d %05d\n",del[i].first,del[i].second,del[i+1].first);
            
        }
        printf("%05d %d -1\n",del[del.size()-1].first,del[del.size()-1].second);
    }
}
相关推荐
秋说22 分钟前
【PTA数据结构 | C语言版】一元多项式求导
c语言·数据结构·算法
Maybyy35 分钟前
力扣61.旋转链表
算法·leetcode·链表
谭林杰2 小时前
B树和B+树
数据结构·b树
卡卡卡卡罗特3 小时前
每日mysql
数据结构·算法
chao_7893 小时前
二分查找篇——搜索旋转排序数组【LeetCode】一次二分查找
数据结构·python·算法·leetcode·二分查找
蜉蝣之翼❉3 小时前
CRT 不同会导致 fopen 地址不同
c++·mfc
aramae4 小时前
C++ -- STL -- vector
开发语言·c++·笔记·后端·visual studio
lifallen4 小时前
Paimon 原子提交实现
java·大数据·数据结构·数据库·后端·算法
lixzest4 小时前
C++ Lambda 表达式详解
服务器·开发语言·c++·算法
EndingCoder4 小时前
搜索算法在前端的实践
前端·算法·性能优化·状态模式·搜索算法