删除C++ STL列表中的元素

如何在C++ STL列表中插入元素?

本文介绍了STL列表中的删除方面。

1. 使用list :: erase() ：此函数的目的是从列表中删除元素。使用此函数可以删除范围内的单个或多个连续元素。此函数接受2个参数，即开始迭代器和结束迭代器。
   时间复杂度： O(n)其中(n是列表的大小)。

   // C++ code to demonstrate the working of erase()
     
   #include
   #include // for list operations
   using namespace std;
     
   int main()
   {
       // initializing list of integers
       list list1={10,15,20,25,30,35};
         
       // declaring list iterators
       list::iterator it = list1.begin();
       list::iterator it1 = list1.begin();
         
       // incrementing the positions of iterators
       advance(it,2);
       advance(it1,5);
         
       // printing original list
       cout << "The original list is : ";
       for (list::iterator i=list1.begin(); i!=list1.end(); i++)
          cout << *i << " ";
          
       cout << endl;
         
       // using erase() to erase single element
       // erases 20
       list1.erase(it);
         
       // list after deletion 1 element
       cout << "The list after deleting 1 element using erase() : ";
       for (list::iterator i=list1.begin(); i!=list1.end(); i++)
          cout << *i << " ";
          
       cout << endl;
         
       it = list1.begin();
         
       // incrementing the positions of iterators
       advance(it,2);
         
       // using erase() to erase multiple elements
       // erases 25,30
       list1.erase(it,it1);
         
       // list after deletion of multiple elements
       cout << "The list after deleting multiple elements using erase() : ";
       for (list::iterator i=list1.begin(); i!=list1.end(); i++)
          cout << *i << " ";
          
       cout << endl;
     
         
   }
   

   输出：

   The original list is : 10 15 20 25 30 35 
   The list after deleting 1 element using erase() : 10 15 25 30 35 
   The list after deleting multiple elements using erase() : 10 15 35 
   

2. 使用list :: pop_front()和list :: pop_back() ：
   • pop_back() ：此函数从列表中删除最后一个元素。这样可以将列表的大小减少1。
     时间复杂度：O(1)
   • pop_front() ：此函数从列表中删除第一个元素，并移动后续元素。这样可以将列表的大小减少1。
     时间复杂度：O(1)

   // C++ code to demonstrate the working of pop_front()
   // and pop_back()
     
   #include
   #include // for list operations
   using namespace std;
     
   int main()
   {
       // initializing list of integers
       list list1={10,15,20,25,30,35};
         
       // printing original list
       cout << "The original list is : ";
       for (list::iterator i=list1.begin(); i!=list1.end(); i++)
          cout << *i << " ";
          
       cout << endl;
         
       // using pop_front() to erase first element of list
       // pops 10
       list1.pop_front();
         
       // list after deleting first element
       cout << "The list after deleting first element using pop_front() : ";
       for (list::iterator i=list1.begin(); i!=list1.end(); i++)
          cout << *i << " ";
          
       cout << endl;
         
       // using pop_back() to erase last element of list
       // pops 35
       list1.pop_back();
         
       // list after deleting last element
       cout << "The list after deleting last element using pop_back() : ";
       for (list::iterator i=list1.begin(); i!=list1.end(); i++)
          cout << *i << " ";
          
       cout << endl;
         
   }
   

   输出：

   The original list is : 10 15 20 25 30 35 
   The list after deleting first element using pop_front() : 15 20 25 30 35 
   The list after deleting last element using pop_back() : 15 20 25 30 
   

3. 使用remove()和remove_if() ：
   • remove() ：此函数删除在其参数中传递的所有出现的值。它与“ erase()”的不同之处在于“ erase()”按位置删除值，而“ remove()”则删除传递的值。列表的大小通过删除的出现次数而减少。
     时间复杂度： O(n)
   • remove_if() ：此函数删除将“ true”返回给在其参数中传递的函数的值的出现。
     时间复杂度： O(n)

   // C++ code to demonstrate the working of remove()
   // remove_if()
     
   #include
   #include // for list operations
   using namespace std;
     
   // function to pass in argument of "remove_if()"
   bool is_div_5(const int& num) { return num%5==0;} 
     
   int main()
   {
       // initializing list of integers
       list list1={10,14,20,22,30,33,22};
         
       // printing original list
       cout << "The original list is : ";
       for (list::iterator i=list1.begin(); i!=list1.end(); i++)
          cout << *i << " ";
          
       cout << endl;
         
       // using remove() to delete all occurrences of 22
       list1.remove(22);
         
       // list after deleting all 22 occurrences
       cout << "The list after deleting all 22 occurrences : ";
       for (list::iterator i=list1.begin(); i!=list1.end(); i++)
          cout << *i << " ";
          
       cout << endl;
         
       // using remove_if() to delete multiple of 5 
       list1.remove_if(is_div_5);
         
       // list after deleting all multiples of 5
       cout << "The list after deleting all multiples of 5 : ";
       for (list::iterator i=list1.begin(); i!=list1.end(); i++)
          cout << *i << " ";
          
       cout << endl;
         
   }
   

   输出：

   The original list is : 10 14 20 22 30 33 22 
   The list after deleting all 22 occurrences : 10 14 20 30 33 
   The list after deleting all multiples of 5 : 14 33 
   

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