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You’re typing a long text with a broken keyboard. Well it’s not so badly broken. The only problem with the keyboard is that sometimes the “home” key or the “end” key gets automatically pressed (internally).

You’re not aware of this issue, since you’re focusing on the text and did not even turn on the monitor! After you finished typing, you can see a text on the screen (if you turn on the monitor).

In Chinese, we can call it Beiju. Your task is to find the Beiju text.

Input

There are several test cases. Each test case is a single line containing at least one and at most 100,000 letters, underscores and two special characters ‘[’ and ‘]’. ‘[’ means the “Home” key is pressed internally, and ‘]’ means the “End” key is pressed internally. The input is terminated by end-of-file (EOF).

Output

For each case, print the Beiju text on the screen.

Sample Input

This_is_a_[Beiju]_text

[[]][][]Happy_Birthday_to_Tsinghua_University Sample

Output

BeijuThis_is_a__text

Happy_Birthday_to_Tsinghua_University

题目意思:

进行文本的输入,如果遇到'['就把光标移动到最前面,遇到']'就把光标移到最后面

解法:

用链表的方法,储存每个字符的位置在输出,这个方法还不熟悉,所以找了新的方法

#include <iostream>
#include <string.h>

using namespace std;

const int MAX = 100000 + 2000;

int main()
{
    char a[MAX];

    while(cin>>(a+1))
    {
        int len = strlen(a+1);
        int next[MAX];
        int last = 0,cur = 0;
        next[0] = 0;
        for(int i = 1;i <=len;i++)
        {
            char ch = a[i];
            if( ch == '[') cur = 0;
            else if(ch == ']') cur = last;
            else {
                next[i] = next[cur];
               // cout<<"i  == cur == netx[i] == netx[cur]="<<i<<cur<<next[i]<<next[cur]<<endl;

                next[cur] = i;
               // cout<<"cur **==netx[cur] **== "<<cur<<"   "<<next[cur]<<endl;

                if(cur == last) last = i;
                cur = i;
               //cout<<"cur last =="<<cur<<last<<endl;
            }
        }
        for(int i = next[0];i!=0;i =next[i])
                cout<<a[i];
        cout<<endl;
    }

    return 0;
}

 

解法2:双向队列<duque>,一种在前后都可以进出元素的队列;这是在网上看到的方法,还有在字符串处理上是我没有想到的;

#include <iostream>
#include <queue>
#include <string.h>

using namespace std;



int main()
{
    char a[100000];
    while(cin>>a)
    {

        char b[100000];
        int bi = 0;
        deque<char> d;
        deque<char>::iterator it;
        int len = strlen(a);
        int ti =1;
        int i = 0;
        while(1)
        {
            if(i>=len)
                break;
            bi = 0;
            while(i<len&&a[i]!='['&&a[i]!=']')
            {
                b[bi++] = a[i++];
            }
            if(ti == 1)
                d.insert(d.begin(),b,b+bi);
            if(ti ==-1)
                d.insert(d.end(),b,b+bi);
            if(a[i]=='[')
                ti =1,i++;
            if(a[i]==']')
                ti =-1,i++;

        }



        for(it=d.begin();it!=d.end();it++)
        {
            cout << *it;
        }
        cout<<endl;


    }
    return 0;
}