[12. maximum continuous non repeating subsequence]

Two cases of double pointer algorithm

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

using namespace std;

int main()
{
    
    char str [1000];
   //scanf("%s", str);
   
   
     fgets(str,sizeof(str), stdin);
     int n = strlen(str);
    for (int i = 0; i < n; i ++)
    {
    
        int j = i;
        while (j < n && str[j] != ' ') j ++;
        
        //  The specific logic of this problem 
        for (int k = i; k < j; k ++) cout << str[k];
        cout << endl;
        
        i = j;
    }
    return 0;
}

 Running results :
 Input ：
abc def cde
 Output ：
abc
def
cde

Maximum continuous unrepeated subsequence

• Open up an additional array S[N], Record dynamically , How many times does the element appear , amount to i Move back one space at a time , Just go S[N] Add an element to , If j Move one space forward , It is equivalent to repeated numbers , In from S[N] Delete the element from .
• Finally, we can make dynamic statistics , How many numbers are there in this array

Given a length of n Integer sequence of , Please find the longest continuous interval that does not contain repeated numbers , Output its length .

subject

Given a length of n Integer sequence of , Please find the longest continuous interval that does not contain repeated numbers , Output its length .

Input format

The first line contains integers n.

The second line contains n It's an integer （ Both in 0∼100000 Within the scope of ）, Represents a sequence of integers .

Output format

All in one line , Contains an integer , Represents the length of the longest continuous interval without repetition .

Data range

1 ≤ n ≤ 100000

sample input ：

5
1 2 2 3 5

sample output ：

3

Code

#include<iostream>
using namespace std;

const int N = 100010;
int n;
int a[N],s[N];
int main()
{
     

   cin >> n;
   for (int i = 0; i < n; i ++)  cin >> a[i];
   
   int res = 0;
   for (int i = 0, j = 0; i < n; i ++)
   {
     
       s[a[i]] ++;
       while (s[a[i]] > 1)
       {
     
           s[a[j]] --;
           j ++;
       }
       res = max(res, i - j + 1);
   }
   cout << res << endl;
   return 0;
}