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 list queue

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[128] = {0};
        char ch;
        int i = 0;
        for(i = 0; 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]] == 1)
            {
                if(tmpch.empty())
                {
                   result.push_back(str[i]);
                }else
                {
                    result.push_back(tmpch[0]);
                }
                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('0');
                }else
                {
                   result.push_back(tmpch[0]);
                }
            }
            //debug(result);
        }
        for(i = 0; i < N; i++)
        {
           if(result[i] == '0') cout << "-1 ";
           else cout << result[i] << ' ';
        }
        cout << endl;
    }

    return 0;
}

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[26];
      queue<char> q;
      cin>>n;
      char s[n];
      for(i=0;i<n;i++)
        cin>>s[i];
      for(i=0;i<26;i++)
        arr[i]=0;
      for(i=0;i<n;i++)
      {
          if(arr[s[i]-'a']==0)
          {
          q.push(s[i]);
          arr[s[i]-'a']=1;
          }
          else
            arr[s[i]-'a']+=1;

          while(!q.empty()&&arr[q.front()-'a']!=1)
            q.pop();
          if(q.empty())
            cout<<-1<<" ";
           else
            cout<<q.front()<<" ";
      }
      cout<<endl;
    }
    return 0;
}

 

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(26,0);
        queue<char> q;
        cin>>n;
        FOR(i,0,n)
        {
            cin>>c;
            v[c-'a']++;
            if(v[c-'a']<=1)
            {
                q.push(c);
            }
            char ans;
            while(!q.empty())
            {
                if(v[q.front()-'a']<=1)
                {
                    ans=q.front();
                    break;
                }
                else
                {
                    q.pop();
                }
            }
            if(q.empty())
            {
                cout<<-1<<' ';
            }
            else
            {
                cout<<ans<<' ';
            }
        }
        cout<<endl;
    }
    
    return 0;
}

 0.115s