极客笔记C++ 算法 rotate()函数
C++算法 rotate() 函数用于旋转范围[first, last)内的元素的顺序。
- 序列将以源序列中间的元素为起点，最后一个元素将跟在第一个元素后面。
- 中间到中间和最后一个元素之间的元素。
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
参数
first ：一个前向迭代器，指向要旋转的范围中第一个元素的位置。
middle ：一个前向迭代器，指向范围[first，last)中要移动到范围中的第一个位置的元素。
last ：一个前向迭代器，指向范围中最后一个元素的位置之后的位置，即将反转元素的范围的末尾。
返回值
无
复杂度
复杂度在范围[first，last)是线性的 ：交换或移动元素，直到所有元素都被重新定位。
数据竞争
范围[first，last)中的对象被修改。
异常
如果元素交换、移动或迭代器操作抛出异常，此函数会抛出异常。
请注意，无效的参数会导致未定义的行为。
示例1
让我们看一个简单的示例，旋转给定的字符串：
#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
示例2
让我们看一个简单的示例：
#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
示例3
让我们看另一个简单的示例：
#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
示例4
再来看一个简单的示例：
#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 ).