用于检查字符串是否可以通过将另一个字符串旋转 d 位置来获得的 Python3 程序

给定两个字符串str1和str2以及一个整数d ，任务是检查str2是否可以通过将str1旋转d位（向左或向右）获得。

例子：

Input: str1 = “abcdefg”, str2 = “cdefgab”, d = 2 
Output: Yes 
Rotate str1 2 places to the left.

Input: str1 = “abcdefg”, str2 = “cdfdawb”, d = 6 
Output: No 
 

方法：这里讨论了解决相同问题的方法。在本文中，反转算法用于在 O(n) 内将字符串向左和向右旋转。如果str1的任何一个旋转等于str2则打印Yes否则打印No 。

下面是上述方法的实现：

Python3

# Python3 implementation of the approach
 
# Function to reverse an array from left
# index to right index (both inclusive)
def ReverseArray(arr, left, right) :
     
    while (left < right) :
        temp = arr[left];
        arr[left] = arr[right];
        arr[right] = temp;
        left += 1;
        right -= 1;
 
# Function that returns true if str1 can be
# made equal to str2 by rotating either
# d places to the left or to the right
def RotateAndCheck(str1, str2, d) :
 
    if (len(str1) != len(str2)) :
        return False;
 
    # Left Rotation string will contain
    # the string rotated Anti-Clockwise
    # Right Rotation string will contain
    # the string rotated Clockwise
    left_rot_str1 = []; right_rot_str1 = [];
    left_flag = True; right_flag = True;
    str1_size = len(str1);
 
    # Copying the str1 string to left rotation string
    # and right rotation string
    for i in range(str1_size) :
        left_rot_str1.append(str1[i]);
        right_rot_str1.append(str1[i]);
 
    # Rotating the string d positions to the left
    ReverseArray(left_rot_str1, 0, d - 1);
    ReverseArray(left_rot_str1, d, str1_size - 1);
    ReverseArray(left_rot_str1, 0, str1_size - 1);
 
    # Rotating the string d positions to the right
    ReverseArray(right_rot_str1, 0, str1_size - d - 1);
    ReverseArray(right_rot_str1,
                 str1_size - d, str1_size - 1);
    ReverseArray(right_rot_str1, 0, str1_size - 1);
 
    # Comparing the rotated strings
    for i in range(str1_size) :
 
        # If cannot be made equal with left rotation
        if (left_rot_str1[i] != str2[i]) :
            left_flag = False;
 
        # If cannot be made equal with right rotation
        if (right_rot_str1[i] != str2[i]) :
            right_flag = False;
 
    # If both or any one of the rotations
    # of str1 were equal to str2
    if (left_flag or right_flag) :
        return True;
    return False;
 
# Driver code
if __name__ == "__main__" :
 
    str1 = list("abcdefg");
    str2 = list("cdefgab");
 
    # d is the rotating factor
    d = 2;
 
    # In case length of str1 < d
    d = d % len(str1);
 
    if (RotateAndCheck(str1, str2, d)) :
        print("Yes");
    else :
        print("No");
 
# This code is contributed by AnkitRai01

输出：

Yes

时间复杂度： O(n)

请参阅完整文章检查是否可以通过旋转另一个字符串d 位置获得字符串以获取更多详细信息！