PAT 1097. Deduplication on a Linked List (链表）

最新推荐文章于 2025-08-13 10:25:05 发布

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最新推荐文章于 2025-08-13 10:25:05 发布

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文章标签：数据结构与算法

原文链接：http://www.cnblogs.com/A-Little-Nut/p/8421565.html

本文介绍了一个算法问题，即如何从一个单链表中移除具有相同绝对值的所有重复节点，并保持这些被移除节点的顺序。文章详细解释了输入输出格式，并提供了一段C++代码实现。

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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:

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:

00100 5
99999 -7 87654
23854 -15 00000
87654 15 -1
00000 -15 99999
00100 21 23854

Sample Output:

00100 21 23854
23854 -15 99999
99999 -7 -1
00000 -15 87654
87654 15 -1

#include<iostream>
#include<math.h>
#include<vector>
using namespace std;
struct node{
    int id,data,next;
};
int main(){
    int s,n,b=-1,index,check[10000]={0},t1=-1,last=-1,tag=0;
    cin>>s>>n;
    node list[100000];
    for(int i=0;i<n;i++){
        node nod;
        cin>>nod.id>>nod.data>>nod.next;
        list[nod.id]=nod;
    }
    index=s;
    vector<node> single,del;
    while(index!=-1){
          if(check[abs(list[index].data)]==0)
             single.push_back(list[index]);
          else
             del.push_back(list[index]);
          check[abs(list[index].data)]=1;
          index=list[index].next;
    }
    for(int i=0;i<single.size();i++){
        if(i!=single.size()-1)
           printf("%05d %d %05d\n",single[i].id,single[i].data,single[i+1].id);
        else
           printf("%05d %d %d\n",single[i].id,single[i].data,-1);
    }
    for(int i=0;i<del.size();i++){
        if(i!=del.size()-1)
           printf("%05d %d %05d\n",del[i].id,del[i].data,del[i+1].id);
        else
           printf("%05d %d %d\n",del[i].id,del[i].data,-1);
    }
    return 0;
}

转载于:https://www.cnblogs.com/A-Little-Nut/p/8421565.html