First non-repeating character in a stream

First non-repeating character in a stream

Given an input stream of n characters consisting only of small case alphabets the task is to find the first non repeating character each time a character is inserted to the stream.

Example

Flow in stream : a, a, b, c
a goes to stream : 1st non repeating element a (a)
a goes to stream : no non repeating element -1 (5, 15)
b goes to stream : 1st non repeating element is b (a, a, b)
c goes to stream : 1st non repeating element is b (a, a, b, c)

Input:
The first line of input contains an integer T denoting the no of test cases. Then T test cases follow. Each test case contains an integer N denoting the size of the stream. Then in the next line are x characters which are inserted to the stream.

Output:
For each test case in a new line print the first non repeating elements separated by spaces present in the stream at every instinct when a character is added to the stream, if no such element is present print -1.

Constraints:
1<=T<=200
1<=N<=500

Example:
Input:
2
4
a a b c
3
a a c 
Output:
a -1 b b
a -1 c

如何处理重复是个难点,本题考察的是队列

hashing  linked listqueue

纯字符串处理方法.

 #include <algorithm>
 #include <iostream>
 #include <string>

 using namespace std;

 int main() {//by guxuanqing@gmai.com
     int T,N;
     cin >> T;
     //getchar();
     string str;
     string result;
     string tmpch;
     while(T--)
     {
         cin >> N;//debug(N);
         //getchar();
         str.clear();
         result.clear();
         tmpch.clear();
         int hashs[] = {};
         char ch;
         int i = ;
         for(i = ; i < N; i++)
         {
             cin >> ch;//debug(ch);
             while (ch == ' ') {
                 cin >> ch;
             }
             //getchar();
             str.push_back(ch);//debug(str);
             ++hashs[(int)str[i]]; //debug(hashs[str[i]]);
             if(hashs[(int)str[i]] == )
             {
                 if(tmpch.empty())
                 {
                    result.push_back(str[i]);
                 }else
                 {
                     result.push_back(tmpch[]);
                 }
                 tmpch.push_back(str[i]);//the first time occurs,push it back to tmpch
             }else
             {
                 // string::size_type n = tmpch.find(str[i]);debug(tmpch);
                 // if(n != string::npos)
                 // {
                 //     tmpch.erase(n,n);debug(tmpch);
                 // }
                 auto n = std::find(tmpch.begin(), tmpch.end(), str[i]);
                 if(n != tmpch.end()) tmpch.erase(n);
                 if(tmpch.empty())
                 {
                     result.push_back('');
                 }else
                 {
                    result.push_back(tmpch[]);
                 }
             }
             //debug(result);
         }
         for(i = ; i < N; i++)
         {
            if(result[i] == '') cout << "-1 ";
            else cout << result[i] << ' ';
         }
         cout << endl;
     }

     return ;
 }

0.093s

 

栈处理的方法:

 //by Amit Negi 2
 #include <iostream>
 #include<queue>
 using namespace std;

 int main() {
     // your code goes here
     int t;
     cin>>t;
     while(t--)
     {
       int n,i,j;
       int arr[];
       queue<char> q;
       cin>>n;
       char s[n];
       for(i=;i<n;i++)
         cin>>s[i];
       for(i=;i<;i++)
         arr[i]=;
       for(i=;i<n;i++)
       {
           if(arr[s[i]-'a']==)
           {
           q.push(s[i]);
           arr[s[i]-'a']=;
           }
           else
             arr[s[i]-'a']+=;

           while(!q.empty()&&arr[q.front()-'a']!=)
             q.pop();
           if(q.empty())
             cout<<-<<" ";
            else
             cout<<q.front()<<" ";
       }
       cout<<endl;
     }
     return ;
 }

0.082s

 //by Ayush Bansal 9
 #include <bits/stdc++.h>
 #define FOR(i,a,b) for(int i=a;i<b;i++)
 using namespace std;

 int main() {
     //code
     int t,n;
     char c;
     cin>>t;
     while(t--)
     {
         vector<int> v(,);
         queue<char> q;
         cin>>n;
         FOR(i,,n)
         {
             cin>>c;
             v[c-'a']++;
             if(v[c-'a']<=)
             {
                 q.push(c);
             }
             char ans;
             while(!q.empty())
             {
                 if(v[q.front()-'a']<=)
                 {
                     ans=q.front();
                     break;
                 }
                 else
                 {
                     q.pop();
                 }
             }
             if(q.empty())
             {
                 cout<<-<<' ';
             }
             else
             {
                 cout<<ans<<' ';
             }
         }
         cout<<endl;
     }

     return ;
 }

0.115s