原文:
给定一个大小为 n 的arr【】和一个值 k，任务是。
示例:
输入: arr = {1，3，5，7，9}，k = 2 输出: 7 9 1 3 5
输入: arr = {1，2，3，4，5}，k = 4 t3】输出: 2 3 4 5 1
算法:给定的问题可以通过来解决。可以遵循以下步骤来解决问题:
- 将所有数组元素从 1 反转到 n -1
- 将数组元素从 1 反转到k–1
- 将数组元素从 k 反转到 n -1
c
// c implementation for the above approach
#include
using namespace std;
// function to rotate the array
// to the right, k times
void rightrotate(int array[], int n, int k)
{
// reverse all the array elements
reverse(array, array n);
// reverse the first k elements
reverse(array, array k);
// reverse the elements from k
// till the end of the array
reverse(array k, array n);
// print the array after rotation
for (int i = 0; i < n; i ) {
cout << array[i] << " ";
}
cout << endl;
}
// driver code
int main()
{
// initialize the array
int array[] = { 1, 2, 3, 4, 5 };
// find the size of the array
int n = sizeof(array) / sizeof(array[0]);
// initialize k
int k = 4;
// call the function and
// print the answer
rightrotate(array, n, k);
return 0;
}
java 语言(一种计算机语言，尤用于创建网站)
/*package whatever //do not write package name here */
import java.io.*;
class gfg
{
// function to rotate the array
// to the right, k times
static void rightrotate(int[] array, int n, int k)
{
// reverse all the array elements
for (int i = 0; i < n / 2; i ) {
int temp = array[i];
array[i] = array[n - i - 1];
array[n - i - 1] = temp;
}
// reverse the first k elements
for (int i = 0; i < k / 2; i ) {
int temp = array[i];
array[i] = array[k - i - 1];
array[k - i - 1] = temp;
}
// reverse the elements from k
// till the end of the array
for (int i = 0; i < (k n) / 2; i ) {
int temp = array[(i k) % n];
array[(i k) % n] = array[(n - i k - 1) % n];
array[(n - i k - 1) % n] = temp;
}
// print the array after rotation
for (int i = 0; i < n; i ) {
system.out.print(array[i] " ");
}
system.out.println();
}
// driver code
public static void main(string[] args)
{
// initialize the array
int array[] = { 1, 2, 3, 4, 5 };
// find the size of the array
int n = array.length;
// initialize k
int k = 4;
// call the function and
// print the answer
rightrotate(array, n, k);
}
}
// this code is contributed by maddler.
python 3
# python program for the above approach
import math
# function to rotate the array
# to the right, k times
def rightrotate(array, n, k):
# reverse all the array elements
for i in range(math.ceil(n / 2)):
temp = array[i]
array[i] = array[n - i - 1]
array[n - i - 1] = temp
# reverse the first k elements
for i in range(math.ceil(k / 2)):
temp = array[i]
array[i] = array[k - i - 1]
array[k - i - 1] = temp
# reverse the elements from k
# till the end of the array
for i in range(math.ceil((k n) / 2)):
temp = array[(i k) % n]
array[(i k) % n] = array[(n - i k - 1) % n]
array[(n - i k - 1) % n] = temp
# print the array after rotation
for i in range(n):
print(array[i], end=" ")
# driver code
arr = [1, 2, 3, 4, 5]
n = len(arr)
k = 4
# call the function and
# print the answer
rightrotate(arr, n, k)
# this code is contributed by saurabh jaiswal
c
// c# program for the above approach
using system;
class gfg
{
// function to rotate the array
// to the right, k times
static void rightrotate(int []array, int n, int k)
{
// reverse all the array elements
for (int i = 0; i < n / 2; i ) {
int temp = array[i];
array[i] = array[n - i - 1];
array[n - i - 1] = temp;
}
// reverse the first k elements
for (int i = 0; i < k / 2; i ) {
int temp = array[i];
array[i] = array[k - i - 1];
array[k - i - 1] = temp;
}
// reverse the elements from k
// till the end of the array
for (int i = 0; i < (k n) / 2; i ) {
int temp = array[(i k) % n];
array[(i k) % n] = array[(n - i k - 1) % n];
array[(n - i k - 1) % n] = temp;
}
// print the array after rotation
for (int i = 0; i < n; i ) {
console.write(array[i] " ");
}
}
// driver code
public static void main()
{
// initialize the array
int []array = { 1, 2, 3, 4, 5 };
// find the size of the array
int n = array.length;
// initialize k
int k = 4;
// call the function and
// print the answer
rightrotate(array, n, k);
}
}
// this code is contributed by samim hossain mondal.
java 描述语言
output
2 3 4 5 1
时间复杂度 : o(n) 辅助空间 : o(1)
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