C++：两种方法：递归函数 && 简单for循环

解题思路:

注意事项:

参考代码:

//递归函数

#include<iostream>

#include<vector>

using namespace std ;

void ko ( vector<int> v , int &m , int &mlen, int &len , int l )

{

         if ( l == 0 ) 

         {

                 if ( len > mlen )

                 {

                         mlen = len ;

                 }

                 return ;

         }

         if ( m == v[l-1] ) 

         {

                 len ++ ;

         }

         else

         {

                 if ( len > mlen )

                 {

                          mlen = len ;

                 }

                 m = v[l-1] ;

                 len = 1 ;

         }

         return ko(v , m , mlen , len , l-1 ) ;

}

int main ( )

{

         int n = 0 ;

         cin >> n ;

         vector<int> v( n , 0 ) ;

         int len = 0 , mlen = 0 , m = 0 , p = 0 ;

         for ( int i = 0 ; i < n ; i ++ )

         {

                  cin >> v[i] ;

         }

         int l = v.size() ;

         m = v[l-1] ;

         ko ( v , m , mlen, len , l) ;

         cout << mlen << endl ;

         return 0 ;

}

/*

//普通for循环

#include<iostream>

#include<vector>

using namespace std ;

int main ( )

{

        int n = 0 ;

        cin >> n ;

        vector<int> v( n , 0 ) ;

        int len = 0 , mlen = 0 , m = 0 , p = 0 ;

        for ( int i = 0 ; i < n ; i ++ )

        {

                    cin >> v[i] ;

                //这里其实可以不用容器或者数组，随便来个变量放一放就可以了

                //这里我为了方便检查代码错误才存储每一个输入的数字的

                    if ( i == 0 ) m = v[0] ;

                    else

                    {

                            if ( v[i] == m )

                            {

                                        len ++ ;

                            }

                            else

                            {

                                    if ( len > mlen )

                                    {

                                        mlen = len ;

                        //              p = m ;

                                    }

                                    m = v[i] ;

                                    len = 1 ;

                            }          

                    }

        }

        cout << mlen << endl ;

    //  cout << p << endl ;

        return 0 ;

}

*/