关注我们

C++ 算法中的 rotate函数

C++算法 rotate()函数用于在[first，last)参数内旋转元素的顺序。

序列将从源序列中间的元素开始，最后一个元素之后是第一个。
中间和最后一个元素之间的元素。

rotate - 语法

template <class ForwardIterator>
void rotate (ForwardIterator first, ForwardIterator middle,
ForwardIterator last);					//until C++ 11

template <class ForwardIterator>
ForwardIterator rotate (ForwardIterator first, ForwardIterator middle,
ForwardIterator last); 				    //since C++ 11

rotate - 参数

first:一个正向迭代器，指向要旋转参数内第一个元素的位置。

middle:正向迭代器，它寻址到[first，last)参数内的元素，该元素被移至该参数内的第一位置。

last:前向迭代器，在元素反转参数内，将位置指向最后一个元素之后的位置。

rotate - 返回值

没有

rotate - 例子1

让我们看一下旋转给定字符串的简单示例:

在线运行复制代码

#include <algorithm>
#include <iostream>
#include <string>

using namespace std;

int main() {
  string str = "Hello";
  cout << "Before Rotate : "<< str << endl;
  
  rotate(str.begin(), str.begin() + 2, str.end());
  cout <<"After Rotate  : " << str << endl;
  
  return 0;
}

输出:

Before Rotate : Hello
After Rotate  : lloHe

rotate - 例子2

让我们看另一个简单的例子:

在线运行复制代码

#include <iostream>
#include <algorithm>
#include <vector>
#include <iomanip>
using namespace std;
 
void print(char a[], int N)
{   
    for(int i = 0; i < N; i++)
    {
        cout << (i + 1) << ". " << setw(2)
             << left << a[i] << "  ";
    }
    cout << endl;
}
 
int main()
{
    char s[] = {'A', 'B', 'C', 'D', 'E', 'G', 'H'};
    int slen = sizeof(s)/sizeof(char);
 
    cout << "Original order : ";
    print(s, slen);
    cout << "Rotate with \'C\' as middle element" << endl;
    rotate(s, s + 2, s + slen);
    cout << "Rotated order  : ";
    print(s, slen);
    cout << "Rotate with \'G\' as middle element" << endl;
    rotate(s, s + 3, s + slen);
    cout << "Rotated order  : ";
    print(s, slen);
    cout << "Rotate with \'A\' as middle element" << endl;
    rotate(s, s + 3, s + slen);
    cout << "Original order : ";
    print(s, slen);
    
    return 0;
}

输出:

Original order : 1. A   2. B   3. C   4. D   5. E   6. G   7. H   
Rotate with 'C' as middle element
Rotated order  : 1. C   2. D   3. E   4. G   5. H   6. A   7. B   
Rotate with 'G' as middle element
Rotated order  : 1. G   2. H   3. A   4. B   5. C   6. D   7. E   
Rotate with 'A' as middle element
Original order : 1. B   2. C   3. D   4. E   5. G   6. H   7. A

rotate - 例子3

让我们看另一个简单的例子:

复制代码

#include<iostream> 
#include<algorithm> 
#include<vector> 
using namespace std; 

int main () { 
    vector<int> vec1{1,2,3,4,5,6,7,8,9}; 
  
   //Print old vector 
    cout << "Old vector:"; 
    for(int i=0; i < vec1.size(); i++) 
        cout << " " << vec1[i]; 
    cout << "\n"; 
   //Rotate vector left 3 times. 
    int rotL=3; 
  
   //rotate function 
    rotate(vec1.begin(), vec1.begin()+rotL, vec1.end()); 
  
   //Print new vector 
    cout << "New vector after left rotation :"; 
    for (int i=0; i < vec1.size(); i++) 
        cout<<" "<<vec1[i]; 
    cout << "\n\n"; 
  
    vector <int> vec2{1,2,3,4,5,6,7,8,9}; 
  
   //Print old vector 
    cout << "Old vector:"; 
    for (int i=0; i < vec2.size(); i++) 
        cout << " " << vec2[i]; 
    cout << "\n"; 
  
   //Rotate vector right 4 times. 
    int rotR = 4; 
  
   //std::rotate function 
    rotate(vec2.begin(), vec2.begin()+vec2.size()-rotR, vec2.end()); 
  
   //Print new vector 
    cout << "New vector after right rotation :"; 
    for (int i=0; i < vec2.size(); i++) 
        cout << " " << vec2[i]; 
    cout << "\n"; 
  
return 0; 
}

输出:

Old vector : 1 2 3 4 5 6 7 8 9
New vector after left rotation : 4 5 6 7 8 9 1 2 3

Old vector : 1 2 3 4 5 6 7 8 9
New vector after right rotation : 6 7 8 9 1 2 3 4 5

rotate - 例子4

让我们看另一个简单的例子:

在线运行复制代码

#include <vector>  
#include <deque>  
#include <algorithm>  
#include <iostream>  
  
int main( ) {  
   using namespace std;  
   vector <int> v1;  
   deque <int> d1;  
   vector <int>::iterator v1Iter1;  
   deque<int>::iterator d1Iter1;  
  
   int i;  
   for ( i = -3 ; i <= 5 ; i++ )  
   {  
      v1.push_back( i );  
   }  
  
   int ii;  
   for ( ii =0 ; ii <= 5 ; ii++ )  
   {  
      d1.push_back( ii );  
   }  
  
   cout << "Vector v1 is ( " ;  
   for ( v1Iter1 = v1.begin( ) ; v1Iter1 != v1.end( ) ;v1Iter1 ++ )  
      cout << *v1Iter1  << " ";  
   cout << ")." << endl;  
  
   rotate ( v1.begin ( ) , v1.begin ( ) + 3 , v1.end ( ) );  
   cout << "After rotating, vector v1 is ( " ;  
   for ( v1Iter1 = v1.begin( ) ; v1Iter1 != v1.end( ) ;v1Iter1 ++ )  
      cout << *v1Iter1  << " ";  
   cout << ")." << endl;  
  
   cout << "The original deque d1 is ( " ;  
   for ( d1Iter1 = d1.begin( ) ; d1Iter1 != d1.end( ) ;d1Iter1 ++ )  
      cout << *d1Iter1  << " ";  
   cout << ")." << endl;  
  
   int iii = 1;  
   while ( iii <= d1.end ( ) - d1.begin ( ) ) {  
      rotate ( d1.begin ( ) , d1.begin ( ) + 1 , d1.end ( ) );  
      cout << "After the rotation of a single deque element to the back,\n d1 is   ( " ;  
      for ( d1Iter1 = d1.begin( ) ; d1Iter1 != d1.end( ) ;d1Iter1 ++ )  
         cout << *d1Iter1  << " ";  
      cout << ")." << endl;  
      iii++;  
   }  
}

输出:

Vector v1 is ( -3 -2 -1 0 1 2 3 4 5 ).
After rotating, vector v1 is ( 0 1 2 3 4 5 -3 -2 -1 ).
The original deque d1 is ( 0 1 2 3 4 5 ).
After the rotation of a single deque element to the back,
 d1 is   ( 1 2 3 4 5 0 ).
After the rotation of a single deque element to the back,
 d1 is   ( 2 3 4 5 0 1 ).
After the rotation of a single deque element to the back,
 d1 is   ( 3 4 5 0 1 2 ).
After the rotation of a single deque element to the back,
 d1 is   ( 4 5 0 1 2 3 ).
After the rotation of a single deque element to the back,
 d1 is   ( 5 0 1 2 3 4 ).
After the rotation of a single deque element to the back,
 d1 is   ( 0 1 2 3 4 5 ).

祝学习愉快！(内容编辑有误？请选中要编辑内容 -> 右键 -> 修改 -> 提交！)

技术教程推荐

Python核心技术与实战 -〔景霄〕

深入拆解Tomcat & Jetty -〔李号双〕