高级算法设计（数组旋转程序） _数组

本文概述

建议：在继续解决方案之前, 请先在"实践"上解决它。
C ++
C
Java
Python3
C#
的PHP
C ++
C
Java
Python3
C#

写一个rotate(ar[]， d, n)函数，旋转大小为n个元素的数组arr[]。

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将上面的数组旋转2将使数组变成：

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推荐：请在"实践首先, 在继续解决方案之前。方法1(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2(一一旋转)

leftRotate(arr[], d, n) start For i = 0 to i < d Left rotate all elements of arr[] by one end

要旋转一个, 将arr [0]存储在临时变量temp中, 将arr [1]移至arr [0], 将arr [2]移至arr [1]…最后将temp移至arr [n-1]
让我们以相同的示例arr [] = [1、2、3、4、5、6、7], d = 2
将arr []旋转1倍
第一次旋转后得到[2, 3, 4, 5, 5, 6, 7, 1], 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 < bits/stdc++.h> 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[i] = 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 program to rotate an array by // d elements #include < stdio.h> /* 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[i] = 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 elementsclass 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[i] = 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# 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[i] = 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

的PHP

< ?php // PHP program to rotate an array // by d elements/*Function to left Rotate arr[] of size n by 1*/ function leftRotatebyOne(& $arr , $n ) { $temp = $arr [0]; for ( $i = 0; $i < $n - 1; $i ++) $arr [ $i ] = $arr [ $i + 1]; $arr [ $i ] = $temp ; }/*Function to left rotate arr[] of size n by d*/ function leftRotate(& $arr , $d , $n ) { for ( $i = 0; $i < $d ; $i ++) leftRotatebyOne( $arr , $n ); }/* utility function to print an array */ function printArray(& $arr , $n ) { for ( $i = 0; $i < $n ; $i ++) echo $arr [ $i ] . " " ; }// Driver Code $arr = array ( 1, 2, 3, 4, 5, 6, 7 ); $n = sizeof( $arr ); // Function calling leftRotate( $arr , 2, $n ); printArray( $arr , $n ); // This code is contributed // by ChitraNayal ?>

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(一种杂耍算法)
这是方法2的扩展。不要将数组一一移动, 而是将数组分成不同的集合
其中集合数等于n和d的GCD并在集合内移动元素。
如果对于上面的示例数组(n = 7和d = 2), GCD为1, 则元素将仅在一组内移动, 我们仅从temp = arr [0]开始并继续移动arr [I + d]到arr [I], 最后将温度存储在正确的位置。
这是n = 12且d = 3的示例。GCD为3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 < bits/stdc++.h> using namespace std; /*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); }/*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 program to rotate an array by // d elements #include < stdio.h> /* 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# 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

输出：