PAT 1097 Deduplication on a Linked List-Toy模板网

这篇具有很好参考价值的文章主要介绍了PAT 1097 Deduplication on a Linked List。希望对大家有所帮助。如果存在错误或未考虑完全的地方，请大家不吝赐教，您也可以点击"举报违法"按钮提交疑问。

个人学习记录，代码难免不尽人意
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 (≤10 5 ) 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 10 4, 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<cstdio>
#include<iostream>
#include<vector>
#include<algorithm>
#include<string>
#include<cmath>
#include<map>
#include<set>
using namespace std;
struct node{
	int data;
	int address;
	int next; 
}Node[100010];

int main(){
  int first,n,k;
  scanf("%d %d",&first,&n);
  for(int i=0;i<n;i++){
  	 int address,data,next;
  	 scanf("%d%d%d",&address,&data,&next);
  	 node a;
  	 a.address=address;
  	 a.data=data;
  	 a.next=next;
  	 Node[address]=a;
  }
  

  set<int> s;
  s.insert(abs(Node[first].data));
  int now=Node[first].address;
  int newfirst=-1;
  int newtemp=-1;
  bool flag=true;
  while(Node[now].next!=-1){
  	node no=Node[now];
  	if(s.find(abs(Node[no.next].data))==s.end()){
  		s.insert(abs(Node[no.next].data));
  		now=no.next;
	  }
	else{
		Node[now].next=Node[no.next].next;
		if(flag){
			newfirst=no.next;
			newtemp=newfirst;
			flag=false;
		}
		else{
		Node[newtemp].next=no.next;
			newtemp=no.next;
			
		}
	
		
	}
  }
  now=Node[first].address;
  while(now!=-1){
  	if(Node[now].next==-1)
  		printf("%05d %d -1\n",Node[now].address,Node[now].data);
  	else 
	  printf("%05d %d %05d\n",Node[now].address,Node[now].data,Node[now].next);
  	now=Node[now].next;
  }
  if(newfirst!=-1){
  	Node[newtemp].next=-1;
  	int newnow=Node[newfirst].address;
  while(newnow!=-1){
  	if(Node[newnow].next==-1)
  		printf("%05d %d -1\n",Node[newnow].address,Node[newnow].data);
  	else 
	  printf("%05d %d %05d\n",Node[newnow].address,Node[newnow].data,Node[newnow].next);
  	newnow=Node[newnow].next;
  }
  }
  
}

本题真的踩了不少坑，我的做法和《算法笔记》略有不同，其中要注意的地方有1.注意边界，比如当前node的next值为-1，再让node=node.next就很有可能出错，应该事先判断。一定要注意边界值。并且，我的做法还要判断是不是有要删除的节点（即原链表中是否出现了两个绝对值一样的数），如果没有的话有些操作不应该出现。文章来源地址https://www.toymoban.com/news/detail-657266.html

到了这里，关于PAT 1097 Deduplication on a Linked List的文章就介绍完了。如果您还想了解更多内容，请在右上角搜索TOY模板网以前的文章或继续浏览下面的相关文章，希望大家以后多多支持TOY模板网！