数组旋转的Python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。

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Array

将上述阵列旋转2将生成阵列

ArrayRotation1

方法1(使用临时数组) :

Input arr[] = [1, 2, 3, 4, 5, 6, 7], d = 2, n =71) Store 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]

Python3

# function to rotate array by d elements using temp array
def rotateArray(arr, n, d):
temp = []
i = 0
while (i < d):
temp.append(arr[i])
i = i + 1
i = 0
while (d < n):
arr[i] = arr[d]
i = i + 1
d = d + 1
arr[:] = arr[: i] + temp
return arr
# Driver function to test above function
arr = [ 1 , 2 , 3 , 4 , 5 , 6 , 7 ]
print ( "Array after left rotation is: " , end = ' ' )
print (rotateArray(arr, len (arr), 2 ))
# this code is contributed by Anabhra Tyagi


输出

Array after left rotation is:  [3, 4, 5, 6, 7, 1, 2]

时间复杂性: O(n) 辅助空间: O(d)

方法2(逐个旋转) :

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

要按1旋转,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]。

Python3

#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


输出

3 4 5 6 7 1 2 

时间复杂性: O(n*d) 辅助空间: O(1)

方法3(杂耍算法) 这是方法2的扩展。不要逐个移动,而是将数组分成不同的集合 其中集合数等于n和d的GCD,并在集合内移动元素。 如果GCD与上述示例数组(n=7,d=2)一样为1,那么元素将仅在一个集合内移动,我们只需从temp=arr[0]开始,并将arr[I+d]移动到arr[I],最后将temp存储在正确的位置。 下面是一个n=12和d=3的例子。GCD为3分

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

ArrayRotation

          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}

Python3

#Function to left rotate arr[] of size n by d
def leftRotate(arr, d, n):
for i in range (gcd(d,n)):
# 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 = " " )
#Function 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 ]
leftRotate(arr, 2 , 7 )
printArray(arr, 7 )
# This code is contributed by Shreyanshi Arun


输出

3 4 5 6 7 1 2 

时间复杂性: O(n) 辅助空间: O(1)

另一种方法: 使用列表切片

Python3

# Python program using the List
# slicing approach to rotate the array
def rotateList(arr,d,n):
arr[:] = arr[d:n] + arr[ 0 :d]
return arr
# Driver function to test above function
arr = [ 1 , 2 , 3 , 4 , 5 , 6 ]
print (arr)
print ( "Rotated list is" )
print (rotateList(arr, 2 , len (arr)))
# this code is contributed by virusbuddah


输出

[1, 2, 3, 4, 5, 6]Rotated list is[3, 4, 5, 6, 1, 2]

如果阵列需要旋转超过其长度,则应进行mod。

例如:将大小为n的arr[]旋转d,其中d大于n。在这种情况下,应计算d%n,并根据mod后的结果旋转。 请参阅完整的文章 阵列旋转程序 更多细节!

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