用于数组旋转的块交换算法的 C# 程序
编写一个函数 rotate(ar[],d,n),将大小为 n 的 arr[]旋转 d 个元素。
将上面的数组旋转 2 将构成数组
算法:
Initialize A = arr[0..d-1] and B = arr[d..n-1]
1) Do following until size of A is equal to size of B
a) If A is shorter, divide B into Bl and Br such that Br is of same
length as A. Swap A and Br to change ABlBr into BrBlA. Now A
is at its final place, so recur on pieces of B.
b) If A is longer, divide A into Al and Ar such that Al is of same
length as B Swap Al and B to change AlArB into BArAl. Now B
is at its final place, so recur on pieces of A.
2) Finally when A and B are of equal size, block swap them.
递归实现:
C
using System;
class GFG{
// Wrapper over the recursive function
// leftRotateRec()
// It left rotates []arr by d.
public static void leftRotate(int []arr,
int d, int n)
{
leftRotateRec(arr, 0, d, n);
}
public static void leftRotateRec(int []arr, int i,
int d, int n)
{
// Return If number of elements
// to be rotated is zero or equal
// to array size
if(d == 0 || d == n)
return;
// If number of elements to be rotated
// is exactly half of array size
if(n - d == d)
{
swap(arr, i, n - d + i, d);
return;
}
// If A is shorter
if(d < n - d)
{
swap(arr, i, n - d + i, d);
leftRotateRec(arr, i, d, n - d);
}
// If B is shorter
else
{
swap(arr, i, d, n - d);
// This is tricky
leftRotateRec(arr, n - d + i,
2 * d - n, d);
}
}
// UTILITY FUNCTIONS
// Function to print an array
public static void printArray(int []arr,
int size)
{
int i;
for(i = 0; i < size; i++)
Console.Write(arr[i] + " ");
Console.WriteLine();
}
// This function swaps d elements
// starting at index fi with d elements
// starting at index si
public static void swap(int []arr, int fi,
int si, int d)
{
int i, temp;
for(i = 0; i < d; i++)
{
temp = arr[fi + i];
arr[fi + i] = arr[si + i];
arr[si + i] = temp;
}
}
// Driver Code
public static void Main(String[] args)
{
int []arr = { 1, 2, 3, 4, 5, 6, 7 };
leftRotate(arr, 2, 7);
printArray(arr, 7);
}
}
// This code is contributed by amal kumar choubey
输出:
3 5 4 6 7 1 2
迭代实现: 这里是同样算法的迭代实现。这里使用了相同的实用函数 swap()。
C
// C# code for above implementation
static void leftRotate(int []arr, int d, int n)
{
int i, j;
if(d == 0 || d == n)
return;
i = d;
j = n - d;
while (i != j)
{
if(i < j) /*A is shorter*/
{
swap(arr, d-i, d+j-i, i);
j -= i;
}
else /*B is shorter*/
{
swap(arr, d-i, d, j);
i -= j;
}
}
/*Finally, block swap A and B*/
swap(arr, d-i, d, i);
}
// This code is contributed by Rajput-Ji
时间复杂度: O(n) 其他阵轮换法请看以下帖子: https://www.geeksforgeeks.org/array-rotation/ https://www . geeksforgeeks . org/程序换阵-轮换-续-反转-算法/
参考文献: 【http://www.cs.bell-labs.com/cm/cs/pearls/s02b.pdf】 如发现以上程序/算法有任何 bug 或想分享任何关于区块交换算法的补充信息,请写评论。
更多详情请参考数组旋转的块交换算法的完整文章!
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