求从 N 到 M 的最少步数

给定两个整数N和M 。任务是通过执行给定的操作找到从 N 到达 M 的最少步数。

1. 将数字 x 乘以 2。因此，x 变为 2*x。
2. 从数字 x 中减去一。因此，x 变为 x-1。

例子：

Input : N = 4, M = 6
Output : 2
Explanation : Perform operation number 2 on N. 
So, N becomes 3 and then perform operation number 1. 
Then, N becomes 6. So, the minimum number of steps is 2. 

Input : N = 10, M = 1
Output : 9
Explanation : Perform operation number two 
9 times on N. Then N becomes 1.

方法：
这个想法是将问题反转如下：我们应该使用以下操作从 M 开始得到数字 N：

1. 如果是偶数，则将数字除以 2。
2. 在数字上加 1。

现在，最少的操作次数是：

1. 如果 N > M，则返回它们之间的差值，即步数将在 M 上加 1，直到它等于 N。
2. 否则如果 N < M。
   • 继续将 M 除以 2，直到它小于 N。如果 M 是奇数，则先加 1，然后除以 2。一旦 M 小于 N，将它们之间的差与上述操作的计数一起添加到计数中.

下面是上述方法的实现：

C++

// CPP program to find minimum number
// of steps to reach M from N
#include 
using namespace std;
 
// Function to find a minimum number
// of steps to reach M from N
int Minsteps(int n, int m)
{
    int ans = 0;
     
    // Continue till m is greater than n
    while(m > n)
    {
        // If m is odd
        if(m&1)
        {
            // add one
            m++;
            ans++;
        }
         
        // divide m by 2       
        m /= 2;
        ans++;
    }
     
    // Return the required answer
    return ans + n - m;
}
 
// Driver code
int main()
{
    int n = 4, m = 6;
    
    cout << Minsteps(n, m);
     
    return 0;
}

Java

// Java program to find minimum number
// of steps to reach M from N
class CFG
{
     
// Function to find a minimum number
// of steps to reach M from N
static int Minsteps(int n, int m)
{
    int ans = 0;
     
    // Continue till m is greater than n
    while(m > n)
    {
        // If m is odd
        if(m % 2 != 0)
        {
            // add one
            m++;
            ans++;
        }
         
        // divide m by 2    
        m /= 2;
        ans++;
    }
     
    // Return the required answer
    return ans + n - m;
}
 
// Driver code
public static void main(String[] args)
{
    int n = 4, m = 6;
     
    System.out.println(Minsteps(n, m));
}
}
 
// This code is contributed by Code_Mech

Python3

# Python3 program to find minimum number
# of steps to reach M from N
 
# Function to find a minimum number
# of steps to reach M from N
def Minsteps(n, m):
 
    ans = 0
     
    # Continue till m is greater than n
    while(m > n):
 
        # If m is odd
        if(m & 1):
             
            # add one
            m += 1
            ans += 1
         
        # divide m by 2    
        m //= 2
        ans += 1
     
    # Return the required answer
    return ans + n - m
 
# Driver code
n = 4
m = 6
 
print(Minsteps(n, m))
 
# This code is contributed by mohit kumar

C#

// C# program to find minimum number
// of steps to reach M from N
using System;
 
class GFG
{
     
// Function to find a minimum number
// of steps to reach M from N
static int Minsteps(int n, int m)
{
    int ans = 0;
     
    // Continue till m is greater than n
    while(m > n)
    {
        // If m is odd
        if(m % 2 != 0)
        {
            // add one
            m++;
            ans++;
        }
         
        // divide m by 2    
        m /= 2;
        ans++;
    }
     
    // Return the required answer
    return ans + n - m;
}
 
// Driver code
public static void Main()
{
    int n = 4, m = 6;
     
    Console.WriteLine(Minsteps(n, m));
}
}
 
// This code is contributed
// by Akanksha Rai

PHP

 $n)
    {
        // If m is odd
        if($m % 2 != 0)
        {
            // add one
            $m++;
            $ans++;
        }
         
        // divide m by 2    
        $m /= 2;
        $ans++;
    }
     
    // Return the required answer
    return $ans + $n - $m;
}
 
// Driver code
$n = 4; $m = 6;
 
echo(Minsteps($n, $m));
 
// This code is contributed by Code_Mech
?>

Javascript

// JavaScript program to find minimum number
// of steps to reach M from N
 
// Function to find a minimum number
// of steps to reach M from N
function Minsteps(n, m)
{
    let ans = 0;
     
    // Continue till m is greater than n
    while(m > n)
    {
        // If m is odd
        if(m&1)
        {
            // add one
            m++;
            ans++;
        }
         
        // divide m by 2       
        m = Math.floor(m / 2);
        ans++;
    }
     
    // Return the required answer
    return ans + n - m;
}
 
// Driver code
    let n = 4, m = 6;   
    document.write(Minsteps(n, m));
 
// This code is contributed by Surbhi Tyagi.

输出：

2

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