获得相同字符串所需的最小旋转的 Python3 程序
给定一个字符串,我们需要找到获得相同字符串所需的最小旋转次数。
例子:
Input : s = "geeks"
Output : 5
Input : s = "aaaa"
Output : 1
这个想法是基于下面的帖子。
检查字符串是否相互旋转的程序
第 1 步:初始化结果 = 0(这里的结果是旋转计数)
第 2 步:取一个与原始字符串相等的临时字符串与其自身连接。
第 3 步:现在从第二个字符(或索引 1)开始取大小与原始字符串相同的临时字符串的子字符串。
第 4 步:增加计数。
步骤 5:检查子字符串是否与原始字符串相等。如果是,则打破循环。否则转到第 2 步并从下一个索引开始重复。
Python3
# Python 3 program to determine minimum
# number of rotations required to yield
# same string.
# Returns count of rotations to get the
# same string back.
def findRotations(str):
# tmp is the concatenated string.
tmp = str + str
n = len(str)
for i in range(1, n + 1):
# substring from i index of
# original string size.
substring = tmp[i: i+n]
# if substring matches with
# original string then we will
# come out of the loop.
if (str == substring):
return i
return n
# Driver code
if __name__ == '__main__':
str = "abc"
print(findRotations(str))
# This code is contributed
# by 29AjayKumar.
Python3
# Python 3 program to determine minimum
# number of rotations required to yield
# same string.
# input
string = 'aaaa'
check = ''
for r in range(1, len(string)+1):
# checking the input after each rotation
check = string[r:] + string[:r]
# following if statement checks if input is
# equals to check , if yes it will print r and
# break out of the loop
if check == string:
print(r)
break
# This code is contributed
# by nagasowmyanarayanan.
输出:
3
时间复杂度: O(n 2 )
Python中的替代实现:
Python3
# Python 3 program to determine minimum
# number of rotations required to yield
# same string.
# input
string = 'aaaa'
check = ''
for r in range(1, len(string)+1):
# checking the input after each rotation
check = string[r:] + string[:r]
# following if statement checks if input is
# equals to check , if yes it will print r and
# break out of the loop
if check == string:
print(r)
break
# This code is contributed
# by nagasowmyanarayanan.
输出:
1
有关更多详细信息,请参阅有关获得相同字符串所需的最小旋转的完整文章!