以2的幂次幂跃升到第N楼梯的最小步伐

给定N个楼梯，任务是找到达到第N个楼梯所需的2次幂的最小跳跃数。

例子：

Input: N = 5 
Output: 
Explanation: 
We can take jumps from 0->1->4. 
So the minimum jumps require are 2.

Input: N = 23 
Output: 4 
Explanation: 
We can take jumps from 0->1->2->4->16. 
So the minimum jumps required are 4. 
 

方法：由于要求跳数必须为2的整数幂，因此给定数N中的设置位的计数是达到第N个楼梯所需的最小跳数，因为所有设置位索引i的2 ^i之和为等于N。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include "bits/stdc++.h"
using namespace std;
 
// Function to count the number of jumps
// required to reach Nth stairs.
int stepRequired(int N)
{
 
    int cnt = 0;
 
    // Till N becomes 0
    while (N) {
 
        // Removes the set bits from
        // the right to left
        N = N & (N - 1);
        cnt++;
    }
 
    return cnt;
}
 
// Driver Code
int main()
{
 
    // Number of stairs
    int N = 23;
 
    // Function Call
    cout << stepRequired(N);
    return 0;
}

Java

// Java program for the above approach
 
import java.util.*;
 
class GFG{
 
// Function to count the number of jumps
// required to reach Nth stairs.
static int stepRequired(int N)
{
 
    int cnt = 0;
 
    // Till N becomes 0
    while (N > 0) {
 
        // Removes the set bits from
        // the right to left
        N = N & (N - 1);
        cnt++;
    }
 
    return cnt;
}
 
// Driver Code
public static void main(String[] args)
{
 
    // Number of stairs
    int N = 23;
 
    // Function Call
    System.out.print(stepRequired(N));
}
}
 
// This code is contributed by PrinciRaj1992

Python3

# Python3 program for the above approach
 
# Function to count the number of jumps
# required to reach Nth stairs.
def stepRequired(N):
 
    cnt = 0;
 
    # Till N becomes 0
    while (N > 0):
 
        # Removes the set bits from
        # the right to left
        N = N & (N - 1);
        cnt += 1;
    return cnt;
 
# Driver Code
if __name__ == '__main__':
 
    # Number of stairs
    N = 23;
 
    # Function Call
    print(stepRequired(N));
     
# This code is contributed by 29AjayKumar

C#

// C# program for the above approach
using System;
 
class GFG{
 
// Function to count the number of
// jumps required to reach Nth stairs.
static int stepRequired(int N)
{
    int cnt = 0;
 
    // Till N becomes 0
    while (N > 0)
    {
 
        // Removes the set bits from
        // the right to left
        N = N & (N - 1);
        cnt++;
    }
 
    return cnt;
}
 
// Driver Code
public static void Main(String[] args)
{
 
    // Number of stairs
    int N = 23;
 
    // Function Call
    Console.Write(stepRequired(N));
}
}
 
// This code is contributed by 29AjayKumar

Javascript

 
// JavaScript program for the above approach
 
// Function to count the number of jumps
// required to reach Nth stairs.
function stepRequired(N)
{
    let cnt = 0;
 
    // Till N becomes 0
    while (N)
    {
         
        // Removes the set bits from
        // the right to left
        N = N & (N - 1);
        cnt++;
    }
 
    return cnt;
}
 
// Driver Code
 
// Number of stairs
let N = 23;
 
// Function Call
document.write(stepRequired(N));
 
// This code is contributed by Surbhi Tyagi
 

输出：

4

时间复杂度： O(log N)