编写一个函数rotate(ar[], d, n) 将大小为 n 的 arr[] 旋转 d 个元素。
将上述数组旋转 2 将使数组
方法一(使用临时数组)
Input arr[] = [1, 2, 3, 4, 5, 6, 7], d = 2, n =7
1) Store the first d elements in a temp array
temp[] = [1, 2]
2) Shift rest of the arr[]
arr[] = [3, 4, 5, 6, 7, 6, 7]
3) Store back the d elements
arr[] = [3, 4, 5, 6, 7, 1, 2]
时间复杂度: O(n)
辅助空间: O(d)
方法二(一一轮换)
leftRotate(arr[], d, n)
start
For i = 0 to i < d
Left rotate all elements of arr[] by one
end
要旋转一,将 arr[0] 存储在临时变量 temp 中,将 arr[1] 移动到 arr[0],将 arr[2] 移动到 arr[1] …最后将 temp 移动到 arr[n-1]
让我们以同样的例子 arr[] = [1, 2, 3, 4, 5, 6, 7], d = 2
将 arr[] 旋转 1 次 2 次
我们在第一次旋转后得到 [2, 3, 4, 5, 6, 7, 1] ,在第二次旋转后得到 [ 3, 4, 5, 6, 7, 1, 2] 。
下面是上述方法的实现:
C++
// C++ program to rotate an array by
// d elements
#include
using namespace std;
/*Function to left Rotate arr[] of
size n by 1*/
void leftRotatebyOne(int arr[], int n)
{
int temp = arr[0], i;
for (i = 0; i < n - 1; i++)
arr[i] = arr[i + 1];
arr[n-1] = temp;
}
/*Function to left rotate arr[] of size n by d*/
void leftRotate(int arr[], int d, int n)
{
for (int i = 0; i < d; i++)
leftRotatebyOne(arr, n);
}
/* utility function to print an array */
void printArray(int arr[], int n)
{
for (int i = 0; i < n; i++)
cout << arr[i] << " ";
}
/* Driver program to test above functions */
int main()
{
int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
int n = sizeof(arr) / sizeof(arr[0]);
// Function calling
leftRotate(arr, 2, n);
printArray(arr, n);
return 0;
}
C
// C program to rotate an array by
// d elements
#include
/* Function to left Rotate arr[] of size n by 1*/
void leftRotatebyOne(int arr[], int n);
/*Function to left rotate arr[] of size n by d*/
void leftRotate(int arr[], int d, int n)
{
int i;
for (i = 0; i < d; i++)
leftRotatebyOne(arr, n);
}
void leftRotatebyOne(int arr[], int n)
{
int temp = arr[0], i;
for (i = 0; i < n - 1; i++)
arr[i] = arr[i + 1];
arr[n-1] = temp;
}
/* utility function to print an array */
void printArray(int arr[], int n)
{
int i;
for (i = 0; i < n; i++)
printf("%d ", arr[i]);
}
/* Driver program to test above functions */
int main()
{
int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
leftRotate(arr, 2, 7);
printArray(arr, 7);
return 0;
}
Java
// Java program to rotate an array by
// d elements
class RotateArray {
/*Function to left rotate arr[] of size n by d*/
void leftRotate(int arr[], int d, int n)
{
for (int i = 0; i < d; i++)
leftRotatebyOne(arr, n);
}
void leftRotatebyOne(int arr[], int n)
{
int i, temp;
temp = arr[0];
for (i = 0; i < n - 1; i++)
arr[i] = arr[i + 1];
arr[n-1] = temp;
}
/* utility function to print an array */
void printArray(int arr[], int n)
{
for (int i = 0; i < n; i++)
System.out.print(arr[i] + " ");
}
// Driver program to test above functions
public static void main(String[] args)
{
RotateArray rotate = new RotateArray();
int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
rotate.leftRotate(arr, 2, 7);
rotate.printArray(arr, 7);
}
}
// This code has been contributed by Mayank Jaiswal
Python3
# Python3 program to rotate an array by
# d elements
# Function to left rotate arr[] of size n by d*/
def leftRotate(arr, d, n):
for i in range(d):
leftRotatebyOne(arr, n)
# Function to left Rotate arr[] of size n by 1*/
def leftRotatebyOne(arr, n):
temp = arr[0]
for i in range(n-1):
arr[i] = arr[i + 1]
arr[n-1] = temp
# utility function to print an array */
def printArray(arr, size):
for i in range(size):
print ("% d"% arr[i], end =" ")
# Driver program to test above functions */
arr = [1, 2, 3, 4, 5, 6, 7]
leftRotate(arr, 2, 7)
printArray(arr, 7)
# This code is contributed by Shreyanshi Arun
C#
// C# program for array rotation
using System;
class GFG {
/* Function to left rotate arr[]
of size n by d*/
static void leftRotate(int[] arr, int d,
int n)
{
for (int i = 0; i < d; i++)
leftRotatebyOne(arr, n);
}
static void leftRotatebyOne(int[] arr, int n)
{
int i, temp = arr[0];
for (i = 0; i < n - 1; i++)
arr[i] = arr[i + 1];
arr[n-1] = temp;
}
/* utility function to print an array */
static void printArray(int[] arr, int size)
{
for (int i = 0; i < size; i++)
Console.Write(arr[i] + " ");
}
// Driver code
public static void Main()
{
int[] arr = { 1, 2, 3, 4, 5, 6, 7 };
leftRotate(arr, 2, 7);
printArray(arr, 7);
}
}
// This code is contributed by Sam007
PHP
Javascript
C++
// C++ program to rotate an array by
// d elements
#include
using namespace std;
/*Function to get gcd of a and b*/
int gcd(int a, int b)
{
if (b == 0)
return a;
else
return gcd(b, a % b);
}
/*Function to left rotate arr[] of siz n by d*/
void leftRotate(int arr[], int d, int n)
{
/* To handle if d >= n */
d = d % n;
int g_c_d = gcd(d, n);
for (int i = 0; i < g_c_d; i++) {
/* move i-th values of blocks */
int temp = arr[i];
int j = i;
while (1) {
int k = j + d;
if (k >= n)
k = k - n;
if (k == i)
break;
arr[j] = arr[k];
j = k;
}
arr[j] = temp;
}
}
// Function to print an array
void printArray(int arr[], int size)
{
for (int i = 0; i < size; i++)
cout << arr[i] << " ";
}
/* Driver program to test above functions */
int main()
{
int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
int n = sizeof(arr) / sizeof(arr[0]);
// Function calling
leftRotate(arr, 2, n);
printArray(arr, n);
return 0;
}
C
// C program to rotate an array by
// d elements
#include
/* function to print an array */
void printArray(int arr[], int size);
/*Fuction to get gcd of a and b*/
int gcd(int a, int b);
/*Function to left rotate arr[] of siz n by d*/
void leftRotate(int arr[], int d, int n)
{
int i, j, k, temp;
/* To handle if d >= n */
d = d % n;
int g_c_d = gcd(d, n);
for (i = 0; i < g_c_d; i++) {
/* move i-th values of blocks */
temp = arr[i];
j = i;
while (1) {
k = j + d;
if (k >= n)
k = k - n;
if (k == i)
break;
arr[j] = arr[k];
j = k;
}
arr[j] = temp;
}
}
/*UTILITY FUNCTIONS*/
/* function to print an array */
void printArray(int arr[], int n)
{
int i;
for (i = 0; i < n; i++)
printf("%d ", arr[i]);
}
/*Fuction to get gcd of a and b*/
int gcd(int a, int b)
{
if (b == 0)
return a;
else
return gcd(b, a % b);
}
/* Driver program to test above functions */
int main()
{
int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
leftRotate(arr, 2, 7);
printArray(arr, 7);
getchar();
return 0;
}
Java
// Java program to rotate an array by
// d elements
class RotateArray {
/*Function to left rotate arr[] of siz n by d*/
void leftRotate(int arr[], int d, int n)
{
/* To handle if d >= n */
d = d % n;
int i, j, k, temp;
int g_c_d = gcd(d, n);
for (i = 0; i < g_c_d; i++) {
/* move i-th values of blocks */
temp = arr[i];
j = i;
while (true) {
k = j + d;
if (k >= n)
k = k - n;
if (k == i)
break;
arr[j] = arr[k];
j = k;
}
arr[j] = temp;
}
}
/*UTILITY FUNCTIONS*/
/* function to print an array */
void printArray(int arr[], int size)
{
int i;
for (i = 0; i < size; i++)
System.out.print(arr[i] + " ");
}
/*Fuction to get gcd of a and b*/
int gcd(int a, int b)
{
if (b == 0)
return a;
else
return gcd(b, a % b);
}
// Driver program to test above functions
public static void main(String[] args)
{
RotateArray rotate = new RotateArray();
int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
rotate.leftRotate(arr, 2, 7);
rotate.printArray(arr, 7);
}
}
// This code has been contributed by Mayank Jaiswal
Python3
# Python3 program to rotate an array by
# d elements
# Function to left rotate arr[] of size n by d
def leftRotate(arr, d, n):
d = d % n
g_c_d = gcd(d, n)
for i in range(g_c_d):
# move i-th values of blocks
temp = arr[i]
j = i
while 1:
k = j + d
if k >= n:
k = k - n
if k == i:
break
arr[j] = arr[k]
j = k
arr[j] = temp
# UTILITY FUNCTIONS
# function to print an array
def printArray(arr, size):
for i in range(size):
print ("% d" % arr[i], end =" ")
# Fuction to get gcd of a and b
def gcd(a, b):
if b == 0:
return a;
else:
return gcd(b, a % b)
# Driver program to test above functions
arr = [1, 2, 3, 4, 5, 6, 7]
n = len(arr)
d = 2
leftRotate(arr, d, n)
printArray(arr, n)
# This code is contributed by Shreyanshi Arun
C#
// C# program for array rotation
using System;
class GFG {
/* Function to left rotate arr[]
of size n by d*/
static void leftRotate(int[] arr, int d,
int n)
{
int i, j, k, temp;
/* To handle if d >= n */
d = d % n;
int g_c_d = gcd(d, n);
for (i = 0; i < g_c_d; i++) {
/* move i-th values of blocks */
temp = arr[i];
j = i;
while (true) {
k = j + d;
if (k >= n)
k = k - n;
if (k == i)
break;
arr[j] = arr[k];
j = k;
}
arr[j] = temp;
}
}
/*UTILITY FUNCTIONS*/
/* Function to print an array */
static void printArray(int[] arr, int size)
{
for (int i = 0; i < size; i++)
Console.Write(arr[i] + " ");
}
/* Fuction to get gcd of a and b*/
static int gcd(int a, int b)
{
if (b == 0)
return a;
else
return gcd(b, a % b);
}
// Driver code
public static void Main()
{
int[] arr = { 1, 2, 3, 4, 5, 6, 7 };
leftRotate(arr, 2, 7);
printArray(arr, 7);
}
}
// This code is contributed by Sam007
Javascript
输出 :
3 4 5 6 7 1 2
时间复杂度: O(n * d)
辅助空间: O(1)
方法 3(杂耍算法)
这是方法2的扩展,不是一个一个移动,而是将数组分成不同的集合
其中集合数等于 n 和 d 的 GCD 并移动集合内的元素。
如果 GCD 为 1,如上述示例数组(n = 7 和 d = 2),则元素将仅在一组内移动,我们只需从 temp = arr[0] 开始并继续移动 arr[I+d]到 arr[I] 并最终将 temp 存储在正确的位置。
这是 n = 12 和 d = 3 的示例。 GCD 是 3 并且
Let arr[] be {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
a) Elements are first moved in first set – (See below
diagram for this movement)
arr[] after this step --> {4 2 3 7 5 6 10 8 9 1 11 12}
b) Then in second set.
arr[] after this step --> {4 5 3 7 8 6 10 11 9 1 2 12}
c) Finally in third set.
arr[] after this step --> {4 5 6 7 8 9 10 11 12 1 2 3}
下面是上述方法的实现:
C++
// C++ program to rotate an array by
// d elements
#include
using namespace std;
/*Function to get gcd of a and b*/
int gcd(int a, int b)
{
if (b == 0)
return a;
else
return gcd(b, a % b);
}
/*Function to left rotate arr[] of siz n by d*/
void leftRotate(int arr[], int d, int n)
{
/* To handle if d >= n */
d = d % n;
int g_c_d = gcd(d, n);
for (int i = 0; i < g_c_d; i++) {
/* move i-th values of blocks */
int temp = arr[i];
int j = i;
while (1) {
int k = j + d;
if (k >= n)
k = k - n;
if (k == i)
break;
arr[j] = arr[k];
j = k;
}
arr[j] = temp;
}
}
// Function to print an array
void printArray(int arr[], int size)
{
for (int i = 0; i < size; i++)
cout << arr[i] << " ";
}
/* Driver program to test above functions */
int main()
{
int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
int n = sizeof(arr) / sizeof(arr[0]);
// Function calling
leftRotate(arr, 2, n);
printArray(arr, n);
return 0;
}
C
// C program to rotate an array by
// d elements
#include
/* function to print an array */
void printArray(int arr[], int size);
/*Fuction to get gcd of a and b*/
int gcd(int a, int b);
/*Function to left rotate arr[] of siz n by d*/
void leftRotate(int arr[], int d, int n)
{
int i, j, k, temp;
/* To handle if d >= n */
d = d % n;
int g_c_d = gcd(d, n);
for (i = 0; i < g_c_d; i++) {
/* move i-th values of blocks */
temp = arr[i];
j = i;
while (1) {
k = j + d;
if (k >= n)
k = k - n;
if (k == i)
break;
arr[j] = arr[k];
j = k;
}
arr[j] = temp;
}
}
/*UTILITY FUNCTIONS*/
/* function to print an array */
void printArray(int arr[], int n)
{
int i;
for (i = 0; i < n; i++)
printf("%d ", arr[i]);
}
/*Fuction to get gcd of a and b*/
int gcd(int a, int b)
{
if (b == 0)
return a;
else
return gcd(b, a % b);
}
/* Driver program to test above functions */
int main()
{
int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
leftRotate(arr, 2, 7);
printArray(arr, 7);
getchar();
return 0;
}
Java
// Java program to rotate an array by
// d elements
class RotateArray {
/*Function to left rotate arr[] of siz n by d*/
void leftRotate(int arr[], int d, int n)
{
/* To handle if d >= n */
d = d % n;
int i, j, k, temp;
int g_c_d = gcd(d, n);
for (i = 0; i < g_c_d; i++) {
/* move i-th values of blocks */
temp = arr[i];
j = i;
while (true) {
k = j + d;
if (k >= n)
k = k - n;
if (k == i)
break;
arr[j] = arr[k];
j = k;
}
arr[j] = temp;
}
}
/*UTILITY FUNCTIONS*/
/* function to print an array */
void printArray(int arr[], int size)
{
int i;
for (i = 0; i < size; i++)
System.out.print(arr[i] + " ");
}
/*Fuction to get gcd of a and b*/
int gcd(int a, int b)
{
if (b == 0)
return a;
else
return gcd(b, a % b);
}
// Driver program to test above functions
public static void main(String[] args)
{
RotateArray rotate = new RotateArray();
int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
rotate.leftRotate(arr, 2, 7);
rotate.printArray(arr, 7);
}
}
// This code has been contributed by Mayank Jaiswal
蟒蛇3
# Python3 program to rotate an array by
# d elements
# Function to left rotate arr[] of size n by d
def leftRotate(arr, d, n):
d = d % n
g_c_d = gcd(d, n)
for i in range(g_c_d):
# move i-th values of blocks
temp = arr[i]
j = i
while 1:
k = j + d
if k >= n:
k = k - n
if k == i:
break
arr[j] = arr[k]
j = k
arr[j] = temp
# UTILITY FUNCTIONS
# function to print an array
def printArray(arr, size):
for i in range(size):
print ("% d" % arr[i], end =" ")
# Fuction to get gcd of a and b
def gcd(a, b):
if b == 0:
return a;
else:
return gcd(b, a % b)
# Driver program to test above functions
arr = [1, 2, 3, 4, 5, 6, 7]
n = len(arr)
d = 2
leftRotate(arr, d, n)
printArray(arr, n)
# This code is contributed by Shreyanshi Arun
C#
// C# program for array rotation
using System;
class GFG {
/* Function to left rotate arr[]
of size n by d*/
static void leftRotate(int[] arr, int d,
int n)
{
int i, j, k, temp;
/* To handle if d >= n */
d = d % n;
int g_c_d = gcd(d, n);
for (i = 0; i < g_c_d; i++) {
/* move i-th values of blocks */
temp = arr[i];
j = i;
while (true) {
k = j + d;
if (k >= n)
k = k - n;
if (k == i)
break;
arr[j] = arr[k];
j = k;
}
arr[j] = temp;
}
}
/*UTILITY FUNCTIONS*/
/* Function to print an array */
static void printArray(int[] arr, int size)
{
for (int i = 0; i < size; i++)
Console.Write(arr[i] + " ");
}
/* Fuction to get gcd of a and b*/
static int gcd(int a, int b)
{
if (b == 0)
return a;
else
return gcd(b, a % b);
}
// Driver code
public static void Main()
{
int[] arr = { 1, 2, 3, 4, 5, 6, 7 };
leftRotate(arr, 2, 7);
printArray(arr, 7);
}
}
// This code is contributed by Sam007
Javascript
输出 :
3 4 5 6 7 1 2
时间复杂度: O(n)
辅助空间: O(1)
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