写题遇到的Set误区

C++ set rbegin() 函数用于返回引用集合容器最后一个元素的反向迭代器。

set 的反向迭代器以相反的方向移动并递增它，直到它到达 set 容器的开头(第一个元素)。

用法

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reverse_iterator rbegin();                            //until C++ 11
const_reverse_iterator rbegin() const;                //until C++ 11
      reverse_iterator rbegin() noexcept;              //since C++ 11
const_reverse_iterator rbegin() const noexcept;  //since C++ 11

参数

空

返回值

它返回一个反向迭代器(反向迭代器)，它指向集合的最后一个元素。

复杂度

恒定。

迭代器有效性

没有变化。

数据竞争

非常量和常量版本都不会访问集合，不会修改集合容器。同时访问集合的元素是安全的。

异常安全

此函数从不抛出异常。

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#include <iostream>
#include <set>

using namespace std;

int main ()
{
  set<int> myset= {10,50,30,40,20};
  
  // show content:
  cout<<"Elements are:"<<endl;
  set<int>::reverse_iterator rit;
  for (rit=myset.rbegin(); rit!=myset.rend(); ++rit)
    cout << *rit<< '\n';

  return 0;
}

输出：

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Elements are:
50
40
30
20
10
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#include <iostream>
#include <set>
#include <string>
#include <iterator>

using namespace std;
 
int main() {
 
    // Creating & Initializing a set of String 
    set<string> setEx = {"aaa", "ccc", "ddd", "bbb"};
 
    // Create a set iterator and point to the end of set
    set<string, int>::reverse_iterator it = setEx.rbegin();
 
    // Iterate over the set using Iterator till beginning.
    while (it != setEx.rend()) {
        // Accessing KEY from element pointed by it.
        string word = *it;
  
        cout << word << endl;
 
        // Increment the Iterator to point to next entry
        it++;
    }
    return 0;
}

输出：

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ddd
ccc
bbb
aaa

因为 set 以键的排序顺序存储元素，因此迭代集合将导致上述顺序，即键的排序顺序。

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#include <set>  
#include <iostream>  
  
int main( )  
{  
   using namespace std;     
   set <int> s1;  
   set <int>::iterator s1_Iter;  
   set <int>::reverse_iterator s1_rIter;  
  
   s1.insert( 10 );  
   s1.insert( 20 );  
   s1.insert( 30 );  
  
   s1_rIter = s1.rbegin( );  
   cout << "The first element in the reversed set is "  
        << *s1_rIter << "." << endl;  
  
   // begin can be used to start an iteration   
   // throught a set in a forward order  
   cout << "The set is:";  
   for ( s1_Iter = s1.begin( ) ; s1_Iter != s1.end( ); s1_Iter++ )  
      cout << " " << *s1_Iter;  
   cout << endl;  
  
   // rbegin can be used to start an iteration   
   // throught a set in a reverse order  
   cout << "The reversed set is:";  
   for ( s1_rIter = s1.rbegin( ) ; s1_rIter != s1.rend( ); s1_rIter++ )  
      cout << " " << *s1_rIter;  
   cout << endl;  
  
   // A set element can be erased by dereferencing to its key   
   s1_rIter = s1.rbegin( );  
   s1.erase ( *s1_rIter );  
  
   s1_rIter = s1.rbegin( );  
   cout << "After the erasure, the first element "  
        << "in the reversed set is "<< *s1_rIter << "." << endl;  
        
return 0;        
}

输出：

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The first element in the reversed set is 30.
The set is:10 20 30
The reversed set is:30 20 10
After the erasure, the first element in the reversed set is 20.
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#include <iostream>
#include <string>
#include <set>

using namespace std;

int main ()
{
  set<int> marks = {400, 350, 465, 290, 410};

   cout << "Marks" << '\n';
   cout<<"______________________\n";
   
  set<int>::reverse_iterator rit;
  for (rit=marks.rbegin(); rit!=marks.rend(); ++rit)
    cout << *rit<< '\n';

    auto ite = marks.rbegin();
 
    cout << "\nHighest Marks is:"<< *ite <<" \n";

  return 0;
  }

输出：

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Marks
______________________
465
410
400
350
290

Highest Marks is:465

Set重载小于号

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#include <iostream>
#include <set>
using namespace std;
struct song
{
    int m_id;
    int m_hot;
    song(int id,int hot)
    {

        this->m_id = id;
        this->m_hot = hot;
    }
    bool operator<(const struct song & right)const   //重载<运算符
    {
        if(this->m_id == right.m_id)     //根据id去重
            return false;
        else
        {
            if(this->m_hot != right.m_hot)
            {
                return this->m_hot > right.m_hot;      //降序
            }
            else
            {
                return this->m_id > right.m_id;     
            }
        }
    }
};
void main()
{
    std::set<song> mySet;
    song s1(10,100);
    song s2(20,200);
    song s3(20,300);
    song s4(30,200);
    mySet.insert(s1);    //插入s1
    mySet.insert(s2);    //插入s2
    mySet.insert(s3);    //s3和s2的id相同，不插入
    mySet.insert(s4);    //插入s4
    for(auto it:mySet)
    {
        std::cout<<"id:"<<it.m_id<<",hot:"<<it.m_hot<<std::endl;
    }
    std::cout<<"end"<<std::endl;
};
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结果如下：
id：30，hot ： 200
id：20，hot ： 200
id：10，hot ： 100end