rotate_copy
交换在两个相邻范围的元素位于源范围内并复制该结果为目标范围。
template<class ForwardIterator, class OutputIterator>
OutputIterator rotate_copy(
ForwardIterator _First,
ForwardIterator _Middle,
ForwardIterator _Last,
OutputIterator _Result
);
参数
_First
解决仅向前迭代器的第一个元素的位置将旋转的大小。_Middle
定义在解决第一个元素位置中的第二部分组件将交换与该范围内的第一部分范围内的仅向前迭代器边界。_ Last
解决仅向前的迭代器通过最终元素的位置一将旋转的大小。_Result
解决输出的迭代器第一个元素的位置在目标范围。
返回值
解决输出的迭代器通过最终元素的位置一在目标范围。
备注
引用的范围必须是有效的;所有指针必须dereferenceable，并在该序列中最后位置以访问按增量。
复杂是线性与至多(_Last – _First)交换。
rotate_copy 有两个相关的窗体:
有关这些功能如何的信息的行为，请参见 经过检查的迭代器。
示例
// alg_rotate_copy.cpp
// compile with: /EHsc
#include <vector>
#include <deque>
#include <algorithm>
#include <iostream>
int main() {
using namespace std;
vector <int> v1 , v2 ( 9 );
deque <int> d1 , d2 ( 6 );
vector <int>::iterator v1Iter , v2Iter;
deque<int>::iterator d1Iter , d2Iter;
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 ( v1Iter = v1.begin( ) ; v1Iter != v1.end( ) ;v1Iter ++ )
cout << *v1Iter << " ";
cout << ")." << endl;
rotate_copy ( v1.begin ( ) , v1.begin ( ) + 3 , v1.end ( ) , v2.begin ( ) );
cout << "After rotating, the vector v1 remains unchanged as:\n v1 = ( " ;
for ( v1Iter = v1.begin( ) ; v1Iter != v1.end( ) ;v1Iter ++ )
cout << *v1Iter << " ";
cout << ")." << endl;
cout << "After rotating, the copy of vector v1 in v2 is:\n v2 = ( " ;
for ( v2Iter = v2.begin( ) ; v2Iter != v2.end( ) ;v2Iter ++ )
cout << *v2Iter << " ";
cout << ")." << endl;
cout << "The original deque d1 is ( " ;
for ( d1Iter = d1.begin( ) ; d1Iter != d1.end( ) ;d1Iter ++ )
cout << *d1Iter << " ";
cout << ")." << endl;
int iii = 1;
while ( iii <= d1.end ( ) - d1.begin ( ) )
{
rotate_copy ( d1.begin ( ) , d1.begin ( ) + iii , d1.end ( ) , d2.begin ( ) );
cout << "After the rotation of a single deque element to the back,\n d2 is ( " ;
for ( d2Iter = d2.begin( ) ; d2Iter != d2.end( ) ;d2Iter ++ )
cout << *d2Iter << " ";
cout << ")." << endl;
iii++;
}
}
Output
Vector v1 is ( -3 -2 -1 0 1 2 3 4 5 ).
After rotating, the vector v1 remains unchanged as:
v1 = ( -3 -2 -1 0 1 2 3 4 5 ).
After rotating, the copy of vector v1 in v2 is:
v2 = ( 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,
d2 is ( 1 2 3 4 5 0 ).
After the rotation of a single deque element to the back,
d2 is ( 2 3 4 5 0 1 ).
After the rotation of a single deque element to the back,
d2 is ( 3 4 5 0 1 2 ).
After the rotation of a single deque element to the back,
d2 is ( 4 5 0 1 2 3 ).
After the rotation of a single deque element to the back,
d2 is ( 5 0 1 2 3 4 ).
After the rotation of a single deque element to the back,
d2 is ( 0 1 2 3 4 5 ).
要求
标头: <algorithm>
命名空间: std