• 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 (≤) which is the total number of nodes. The address of a node is a 5-digit nonnegative integer, and NULL is represented by −.

    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 1, 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

    链表题

    #include <iostream>
    #include <vector>
    #include <unordered_map>
    using namespace std;
    struct node {
        int addr, val, next;
    };
    int main() {
        int start;
        int N;
        node tmp;
        vector<node> v, v_del;
        unordered_map<int, node> m;
        cin >> start >> N;
        while(N--) {
            cin >> tmp.addr >> tmp.val >> tmp.next;
            m[tmp.addr] = tmp;
        }
        unordered_map<int, bool> has;
        while(start != -1) {
            if(!has[abs(m[start].val)]) {
                v.push_back(m[start]);
                has[abs(m[start].val)] = true;
            }else v_del.push_back(m[start]);
            start = m[start].next;
        }
        for(int i = 0; i < v.size(); i++)
            if(i == 0) printf("%05d %d ", v[i].addr, v[i].val);
            else printf("%05d
    %05d %d ", v[i].addr, v[i].addr, v[i].val);
        printf("-1
    ");
        if(v_del.size() > 0) {
            for(int i = 0; i < v_del.size(); i++)
                if(i == 0) printf("%05d %d ", v_del[i].addr, v_del[i].val);
                else printf("%05d
    %05d %d ", v_del[i].addr, v_del[i].addr, v_del[i].val);
            printf("-1
    ");
        }
        return 0;
    }
  • 相关阅读:
    【PAT】 B1006 换个格式输出整数
    【PAT】B1014 福尔摩斯的约会
    【PAT】B1005 继续(3n+1)猜想
    【PAT】B1004 成绩排名
    【PAT】B1003 我要通过!
    【PAT】B1002 写出这个数
    【PAT】B1001 害死人不偿命的(3n+1)猜想
    【PAT】A1001A+B Format
    【PAT】B1027 打印沙漏(20 分)
    【PAT】B1032 挖掘机技术哪家强(20 分)
  • 原文地址:https://www.cnblogs.com/littlepage/p/12905628.html
Copyright © 2020-2023  润新知