1097 Deduplication on a Linked List （25 分)

2019-04-16

1097 Deduplication on a Linked List （25 分)

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:

1	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

作者: CHEN, Yue

单位: 浙江大学

时间限制: 400 ms

内存限制: 64 MB

代码长度限制: 16 KB

题目大意

给出一个链表，要求保留每个关键字的绝对值出现第一次的结点。把删除的结点也组成链表。最后输出剩余结点，以及删除结点。

分析

先用一个map存储地址与结点，然后遍历。用两个vector存储剩余的和删除的结点。这里不需要map，使用vector，结点push的顺序即为链表中的顺序。最后输出

代码

#include <iostream>
#include <vector>
#include <stack>
#include <algorithm>
#include <deque>
#include <map>
#include <set>
#include <cstring>
#include <cmath>
#include <sstream>

using namespace std;
int head, n;

struct node {
    int addr;
    int val;
    int next;

    node(int addr, int val, int next) : addr(addr), val(val), next(next) {}

    node() {}

};

map<int, node> orig;
vector<node> rem;
vector<node> rmv;
map<int,int>ocur;
int main() {
    cin >> head >> n;
    int at, vt, next;
    for (int i = 0; i < n; i++) {
        cin >> at >> vt >> next;
        node n(at, vt, next);
        orig[at] = n;
    }
    for(int i=head;i!=-1;i=orig[i].next){
        node now=orig[i];
        int x=abs(now.val);
        ocur[x]++;
        if(ocur[x]>1)
            rmv.push_back(now);
        else
            rem.push_back(now);
    }

    for(int i=0;i<rem.size();i++){
        if(i<rem.size()-1)
            printf("%05d %d %05d\n",rem[i].addr,rem[i].val,rem[i+1].addr);
        else
            printf("%05d %d -1\n",rem[i].addr,rem[i].val);
    }
    for(int i=0;i<rmv.size();i++){
        if(i<rmv.size()-1)
            printf("%05d %d %05d\n",rmv[i].addr,rmv[i].val,rmv[i+1].addr);
        else
            printf("%05d %d -1\n",rmv[i].addr,rmv[i].val);
    }
    return 0;
}