解锁圆形锁的最小旋转次数
原文:https://www . geesforgeks . org/minimum-rotations-unlock-circular-lock/
给你一把锁，它由 n 个不同的圆环组成，每个圆环上连续印着 0-9 个数字。最初，所有的 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旋转每个环任意次数。你必须找到锁环转动的最小次数才能打开锁。
例:
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，我们可以向前或向后两个方向中的任何一个旋转它，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(a-b)，10-abs(a-b)) 作为 a-b 表示正向旋转的次数， 10-abs(a-b) 表示一个环从 a 旋转到 b 的反向旋转次数。此外，我们必须为每个环找到每个数字的最小旋转次数。因此，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，并在最左边的数字结束。
C++
// CPP program for min rotation to unlock
#include <bits/stdc++.h>
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 语言(一种计算机语言，尤用于创建网站)
// 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 */
Python 3
# 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 mits
C
// 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 Mittal
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program for min
// rotation to unlock
// function for min rotation
function minRotation($input,
$unlock_code)
{
$rotation = 0;
$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
$input = 28756;
$unlock_code = 98234;
echo "Minimum Rotation = "
, minRotation($input, $unlock_code);
// This code is contributed by vt_m.
?>
java 描述语言
<script>
// JavaScript program for min rotation to unlock
// function for min rotation
function minRotation(input, unlock_code)
{
let rotation = 0;
let 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 = Math.floor(input / 10);
unlock_code = Math.floor(unlock_code / 10);
}
return rotation;
}
// Driver Code
let input = 28756;
let unlock_code = 98234;
document.write("Minimum Rotation = "+
minRotation(input, unlock_code));
</script>
输出:
Minimum Rotation = 12
本文由Shivam Pradhan(anuj _ charm)供稿。如果你喜欢 GeeksforGeeks 并想投稿，你也可以使用contribute.geeksforgeeks.org写一篇文章或者把你的文章邮寄到 contribute@geeksforgeeks.org。看到你的文章出现在极客博客主页上，帮助其他极客。 如果发现有不正确的地方，或者想分享更多关于上述话题的信息，请写评论。
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