给你一个锁，它由 n 个不同的圆环组成，每个圆环上都印有 0-9 位数字。最初所有 n 环一起显示一个 n 位整数，但只有特定的代码可以打开锁。您可以在任一方向上任意次数地旋转每个环。您必须找到在锁环上完成的最小旋转次数才能打开锁。
例子：
Input : Input = 2345, Unlock code = 5432
Output : Rotations required = 8
Explanation : 1st ring is rotated thrice as 2->3->4->5
2nd ring is rotated once as 3->4
3rd ring is rotated once as 4->3
4th ring is rotated thrice as 5->4->3->2
Input : Input = 1919, Unlock code = 0000
Output : Rotations required = 4
Explanation : 1st ring is rotated once as 1->0
2nd ring is rotated once as 9->0
3rd ring is rotated once as 1->0
4th ring is rotated once as 9->0对于单个环，我们可以将其向前或向后两个方向中的任何一个旋转，如下所示：
- 0->1->2….->9->0
- 9->8->….0->9
但是我们关心所需的最小旋转次数，所以我们应该选择min (abs(ab), 10-abs(ab))因为ab表示正向旋转的次数，而10-abs(ab)表示向后旋转的次数一个从 a 旋转到 b 的环。此外，我们必须为每个数字找到每个环的最小数量。所以从最右边的数字开始，我们可以很容易地找到每个环所需的最小旋转次数，并在最左边的数字处结束。
C++
// CPP program for min rotation to unlock
#include
using namespace std;
// function for min rotation
int minRotation(int input, int unlock_code)
{
int rotation = 0;
int input_digit, code_digit;
// iterate till input and unlock code become 0
while (input || unlock_code) {
// input and unlock last digit as reminder
input_digit = input % 10;
code_digit = unlock_code % 10;
// find min rotation
rotation += min(abs(input_digit - code_digit),
10 - abs(input_digit - code_digit));
// update code and input
input /= 10;
unlock_code /= 10;
}
return rotation;
}
// driver code
int main()
{
int input = 28756;
int unlock_code = 98234;
cout << "Minimum Rotation = "
<< minRotation(input, unlock_code);
return 0;
} Java
// Java program for min rotation to unlock
class GFG
{
// function for min rotation
static int minRotation(int input, int unlock_code)
{
int rotation = 0;
int input_digit, code_digit;
// iterate till input and unlock code become 0
while (input>0 || unlock_code>0) {
// input and unlock last digit as reminder
input_digit = input % 10;
code_digit = unlock_code % 10;
// find min rotation
rotation += Math.min(Math.abs(input_digit
- code_digit), 10 - Math.abs(
input_digit - code_digit));
// update code and input
input /= 10;
unlock_code /= 10;
}
return rotation;
}
// driver code
public static void main (String[] args) {
int input = 28756;
int unlock_code = 98234;
System.out.println("Minimum Rotation = "+
minRotation(input, unlock_code));
}
}
/* This code is contributed by Mr. Somesh Awasthi */Python3
# Python3 program for min rotation to unlock
# function for min rotation
def minRotation(input, unlock_code):
rotation = 0;
# iterate till input and unlock
# code become 0
while (input > 0 or unlock_code > 0):
# input and unlock last digit
# as reminder
input_digit = input % 10;
code_digit = unlock_code % 10;
# find min rotation
rotation += min(abs(input_digit - code_digit),
10 - abs(input_digit - code_digit));
# update code and input
input = int(input / 10);
unlock_code = int(unlock_code / 10);
return rotation;
# Driver Code
input = 28756;
unlock_code = 98234;
print("Minimum Rotation =",
minRotation(input, unlock_code));
# This code is contributed by mitsC#
// C# program for min rotation to unlock
using System;
class GFG {
// function for min rotation
static int minRotation(int input,
int unlock_code)
{
int rotation = 0;
int input_digit, code_digit;
// iterate till input and
// unlock code become 0
while (input > 0 ||
unlock_code > 0)
{
// input and unlock last
// digit as reminder
input_digit = input % 10;
code_digit = unlock_code % 10;
// find min rotation
rotation += Math.Min(Math.Abs(input_digit -
code_digit), 10 - Math.Abs(
input_digit - code_digit));
// update code and input
input /= 10;
unlock_code /= 10;
}
return rotation;
}
// Driver Code
public static void Main ()
{
int input = 28756;
int unlock_code = 98234;
Console.Write("Minimum Rotation = "+
minRotation(input, unlock_code));
}
}
// This code is contributed by Nitin MittalPHP
Javascript
输出：
Minimum Rotation = 12如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。