【PAT甲级】1097 Deduplication on a Linked List (25 point(s))

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 ($\le 10^5$) 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

Method:

模拟链表模板，根据标记数组 vis 分别装入两个向量，更新下标并格式化打印。

Solution:

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

// 链表节点
const int maxn = 1e6+10;

struct node{
    
    
	int address, key, next;
}nod[maxn]; 

// 标记数组
bool vis[maxn] = {
    
    false};

void print(vector<node> list) {
    
    
	int sum = list.size();
	if(sum != 0){
    
    							// 向量不为空时
        for (int i = 0; i < sum-1; i++)		// 更新下标
		    list[i].next = list[i+1].address;
	    list[sum-1].next = -1;
	    
		for (int i = 0; i < sum-1; i++)		// 格式化打印
			printf("%05d %d %05d\n",list[i].address,list[i].key,list[i+1].address);
		printf("%05d %d -1\n",list[sum-1].address,list[sum-1].key);
	}
}
int main(){
    
    
	int head = 0, n = 0;
	cin >> head >> n;
	
	// 初始化
	int address = 0, key = 0, next = 0;
	for (int i = 0; i < n; i++) {
    
    
		cin >> address >> key >> next;
		nod[address].address = address;
		nod[address].key = key;
		nod[address].next = next;
	}	
	
	// 装入向量
	vector<node> list1, list2;
	for (head; head != -1; head = nod[head].next) {
    
    
		if (!vis[ abs( nod[head].key ) ]) {
    
    
			vis[ abs( nod[head].key ) ] = true;
			list1.push_back(nod[head]);
		}
		else {
    
    
			list2.push_back(nod[head]);
		}
	}
    
    print(list1);   print(list2);


	return 0;
}