zoukankan      html  css  js  c++  java
  • 1057. Stack (30)

    /*to solve the problem ,i think we can use stack to maintain the numbers,
    and list to keep it sorted,which is very important to find the Median number.
    */
    #include<stack>
    #include<iostream>
    #include<string>
    #include<vector>
    #include<list>
    #include<algorithm>
    using namespace std;
    
    int main()
    {
        int n;
        stack<int> stak;
        string command;
        int a;
        cin>>n;
        while (n--)
        {
            cin>>command;
            if (command == "Push")
            {
                cin>>a;
                stak.push(a);
            }
            if (command == "Pop")
            {
                if (stak.empty())
                {
                    cout<<"Invalid"<<endl;
                }
                else
                {
                    cout<<stak.top()<<endl;
                    stak.pop();
                }
            }
            if (command == "PeekMedian")
            {
                if (stak.empty())
                {
                    cout<<"Invalid"<<endl;
                    continue;
                }
                vector<int> ls;
                while (!stak.empty())
                {
                    ls.push_back(stak.top());
                    stak.pop();
                }
                //use the unsorted vector to push_back the stack
                for (int i = ls.size()-1; i >=0 ; i—)
                {
                    stak.push(ls[i]);
                }
                sort(ls.begin(),ls.end());
                cout<<ls[ls.size()%2 == 0 ? ls.size()/2-1:(ls.size()+1)/2-1]<<endl;
            }
        }
    }

    At first,i think i could use a temporary stack not only to keep the stack in order but alse to find

    the median one . And it is proofed that it is feasible but not effcient ,as everytime we need

    to sort the stack as long as we want to use the command “PeekMedian”.

    the Time complexity is O(n * log n). then i searched the web. I find someone use the set,

    which is virtually a balanced binary tree.so we could reduced the time complexity down to

    O(log n). Then i rebuild the code according to this suggestion.

    #include <iostream>
    #include <set>
    #include <algorithm>
    #include <stack>
    #include <cstring>
    
    using namespace std;
    
    multiset<int> small, big;
    stack<int> s;
    int mid;
    
    void adjust()
    {
        if (small.size() > big.size() + 1)
        {
            auto it = small.end();
            --it;
            big.insert(*it);
            small.erase(it);
        }
        else if (small.size() < big.size())
        {
            auto it = big.begin();
            small.insert(*it);
            big.erase(it);
        }
        if (s.size() > 0)
        {
            auto it = small.end();
            --it;
            mid = *it;
        }
    }
    
    int main()
    {
        int n;
        char op[15];
        int top, Key;
        scanf("%d", &n);
        while (n--)
        {
            scanf("%s", op);
            if (op[1] == 'o')
            {
                if (s.size() == 0)
                    printf("Invalid
    ");
                else
                {
                    top = s.top();
                    s.pop();
                    printf("%d
    ", top);
                    if (mid >= top)
                    {
                        auto it = small.find(top);
                        small.erase(it);
                    }
                    else
                    {
                        auto it = big.find(top);
                        big.erase(it);
                    }
                    adjust();
                }
            }
            else if (op[1] == 'u')
            {
                scanf("%d", &Key);
                if (s.size() == 0)
                {
                    small.insert(Key);
                    mid = Key;
                }
                else if (Key <= mid)
                    small.insert(Key);
                else
                    big.insert(Key);
                s.push(Key);
                adjust();
            }
            else if (op[1] == 'e')
            {
                if (s.size() == 0)
                    printf("Invalid
    ");
                else
                    printf("%d
    ", mid);
            }
        }
    }
  • 相关阅读:
    delphi利用qdac qworker计划任务
    delphi libssh2 无法立即完成一个非阻止性套接字操作
    线程池底层原理
    【spring源码分析】二、Spring扩展点的归总
    【spring源码分析】一、spring Refresh流程
    Spring中常用的类
    设计模式-proxy
    SpringAOP
    SpringIOC
    难受,nginx worker进程内存持续飘升!
  • 原文地址:https://www.cnblogs.com/maverick-fu/p/3973240.html
Copyright © 2011-2022 走看看