在O(n)时间和O(1)空间中打印数组的左旋转 · GeeksForGeeks 中文系列教程

# 在`O(n)`时间和`O(1)`空间中打印数组的左旋转 > 原文： [https://www.geeksforgeeks.org/print-left-rotation-array/](https://www.geeksforgeeks.org/print-left-rotation-array/) 给定大小为 n 的数组和多个值，我们需要围绕该值左旋转数组。如何快速打印多个左旋？ **示例**： ``` Input : arr[] = {1, 3, 5, 7, 9} k1 = 1 k2 = 3 k3 = 4 k4 = 6 Output : 3 5 7 9 1 7 9 1 3 5 9 1 3 5 7 3 5 7 9 1 Input : arr[] = {1, 3, 5, 7, 9} k1 = 14 Output : 9 1 3 5 7 ``` 我们在下面的文章中讨论了一个解决方案。 [快速找到数组的多个左旋转| 系列 1](https://www.geeksforgeeks.org/quickly-find-multiple-left-rotations-of-an-array/) 上面讨论的解决方案需要额外的空间。在本文中，我们讨论了不需要额外空间的优化解决方案。 ## C++ ```cpp // CPP implementation of left rotation of // an array K number of times #include <bits/stdc++.h> using namespace std; // Function to leftRotate array multiple times void leftRotate(int arr[], int n, int k) { /* To get the starting point of rotated array */ int mod = k % n; // Prints the rotated array from start position for (int i = 0; i < n; i++) cout << (arr[(mod + i) % n]) << " "; cout << "\n"; } // Driver program int main() { int arr[] = { 1, 3, 5, 7, 9 }; int n = sizeof(arr) / sizeof(arr[0]); int k = 2; leftRotate(arr, n, k); k = 3; leftRotate(arr, n, k); k = 4; leftRotate(arr, n, k); return 0; } ``` ## Java ```java // JAVA implementation of left rotation // of an array K number of times import java.util.*; import java.lang.*; import java.io.*; class arr_rot { // Function to leftRotate array multiple // times static void leftRotate(int arr[], int n, int k) { /* To get the starting point of rotated array */ int mod = k % n; // Prints the rotated array from // start position for(int i = 0; i < n; ++i) System.out.print(arr[(i + mod) % n] + " "); System.out.println(); } // Driver program public static void main (String[] args) { int arr[] = { 1, 3, 5, 7, 9 }; int n = arr.length; int k = 2; leftRotate(arr, n, k); k = 3; leftRotate(arr, n, k); k = 4; leftRotate(arr, n, k); } } // This code is contributed by Sanjal ``` ## Python ``` # Python implementation of left rotation of # an array K number of times # Function to leftRotate array multiple times def leftRotate(arr, n, k): # To get the starting point of rotated array mod = k % n s = "" # Prints the rotated array from start position for i in range(n): print str(arr[(mod + i) % n]), print return # Driver program arr = [ 1, 3, 5, 7, 9 ] n = len(arr) k = 2 leftRotate(arr, n, k) k = 3 leftRotate(arr, n, k) k = 4 leftRotate(arr, n, k) #This code is contributed by Sachin Bisht ``` ## C# ```cs // C# implementation of left // rotation of an array K // number of times using System; class GFG { // Function to leftRotate // array multiple times static void leftRotate(int []arr, int n, int k) { // To get the starting // point of rotated array int mod = k % n; // Prints the rotated array // from start position for(int i = 0; i < n; ++i) Console.Write(arr[(i + mod) % n] + " "); Console.WriteLine(); } // Driver Code static public void Main () { int []arr = {1, 3, 5, 7, 9}; int n = arr.Length; int k = 2; leftRotate(arr, n, k); k = 3; leftRotate(arr, n, k); k = 4; leftRotate(arr, n, k); } } // This code is contributed by m_kit ``` ## PHP ```php <?php // PHP implementation of // left rotation of an // array K number of times // Function to leftRotate // array multiple times function leftRotate($arr, $n, $k) { // To get the starting // point of rotated array $mod = $k % $n; // Prints the rotated array // from start position for ($i = 0; $i < $n; $i++) echo ($arr[($mod + $i) % $n]) , " "; echo "\n"; } // Driver Code $arr = array(1, 3, 5, 7, 9); $n = sizeof($arr); $k = 2; leftRotate($arr, $n, $k); $k = 3; leftRotate($arr, $n, $k); $k = 4; leftRotate($arr, $n, $k); // This code is contributed by m_kit ?> ``` **输出**： ``` 5 7 9 1 3 7 9 1 3 5 9 1 3 5 7 ```