ural Cipher Message

Cipher Message

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Description

Müller tried to catch Stierlitz red-handed many times, but always failed because Stierlitz could ever find some excuse. Once Stierlitz was looking through his email messages. At that moment, Müller entered secretly and watched a meaningless sequence of symbols appear on the screen. “A cipher message,” Müller thought. “UTF-8,” Stierlitz thought.

It is known that Stierlitz ciphers messages by the following method.

1. He deletes all spaces and punctuation marks.
2. He replaces all successive identical letters by one such letter.
3. He inserts two identical letters at an arbitrary place many times.

Try to restore a message as it was after the second step. For that, remove from the message all pairs of identical letters inserted at the third step.

Input

The only input line contains a message ciphered by Stierlitz. The message consists of lowercase English letters and its length is at most 200000.

Output

Output the restored message.

Sample Input

input	output
wwstdaadierfflitzzz	stierlitz

 

解答：

首先采用了最直接的做法，这种做法时间复杂度为O(n²)

#include <iostream>
#include <cstring>
using namespace std;

int main(){
    string data;
    cin>>data;
    int a;
    while(1){
        a=0;
        for(int i=0;i<data.length();i++){
            if(data[i]==data[i+1]){
                a=1;
                data.erase(i,2);
            }
        }
        if(a==0){
            break;
        }
    }
    cout<<data;
    
} 

结果超时，于是用下面的方法另起炉灶。时间复杂度为O(n)，遂AC

#include <iostream>
#include <cstring>
using namespace std;
#define N 200002

int main(){
    char  data[N];
    gets(data);
    char temp[N];
    memset(temp,0,sizeof(temp));
    
    int fir=-1;
    int lon=strlen(data);
    for(int i=0;i<lon;i++){
        if(data[i] != temp[fir] ){
            temp[++fir]=data[i];
        }
        else{
            fir--;
        }
    }
    
    for(int i=0;i<=fir;i++){
        cout<<temp[i];
    }
    
    return 0;
}