给定两个数字N和K。任务是从右侧找到数字中第K个设置位的索引。
注意：二进制表示形式的索引从右侧的0开始。例如，在二进制数“ 000011”中，第一个设置位在从右开始的索引0处，第二个设置位在从右开始的索引1处。
例子：
Input: N = 15, K = 3
Output: 2
15 is "1111", hence the third bit is at index 2 from right.
Input: N = 19, K = 2
Output: 1
19 is "10011", hence the second set bit is at index 1 from right. 方法：初始化计数器0，如果数字的最后一位被设置，则将其增加。为了访问下一位，将数字右移1。当计数器的值等于K时，我们返回数字的索引，该索引在每次右移时都会递增。
下面是上述方法的实现：
C++
// C++ program to implement
// the above approach
#include
using namespace std;
// Function that returns the Kth set bit
int FindIndexKthBit(int n, int k)
{
int cnt = 0;
int ind = 0;
// Traverse in the binary
while (n) {
// Check if the last
// bit is set or not
if (n & 1)
cnt++;
// Check if count is equal to k
// then return the index
if (cnt == k)
return ind;
// Increase the index
// as we move right
ind++;
// Right shift the number by 1
n = n >> 1;
}
return -1;
}
// Driver Code
int main()
{
int n = 15, k = 3;
int ans = FindIndexKthBit(n, k);
if (ans != -1)
cout << ans;
else
cout << "No k-th set bit";
return 0;
} Java
// Java program to implement
// the above approach
import java.util.*;
class GFG
{
// Function that returns the Kth set bit
static int FindIndexKthBit(int n, int k)
{
int cnt = 0;
int ind = 0;
// Traverse in the binary
while (n > 0)
{
// Check if the last
// bit is set or not
if ((n & 1 ) != 0)
cnt++;
// Check if count is equal to k
// then return the index
if (cnt == k)
return ind;
// Increase the index
// as we move right
ind++;
// Right shift the number by 1
n = n >> 1;
}
return -1;
}
// Driver Code
public static void main(String args[])
{
int n = 15, k = 3;
int ans = FindIndexKthBit(n, k);
if (ans != -1)
System.out.println(ans);
else
System.out.println("No k-th set bit");
}
}
// This code is contributed by
// Surendra_GangwarPython3
# Python3 implementation of the approach
# Function that returns the Kth set bit
def FindIndexKthBit(n, k):
cnt, ind = 0, 0
# Traverse in the binary
while n > 0:
# Check if the last
# bit is set or not
if n & 1:
cnt += 1
# Check if count is equal to k
# then return the index
if cnt == k:
return ind
# Increase the index
# as we move right
ind += 1
# Right shift the number by 1
n = n >> 1
return -1
# Driver Code
if __name__ == "__main__":
n, k = 15, 3
ans = FindIndexKthBit(n, k)
if ans != -1:
print(ans)
else:
print("No k-th set bit")
# This code is contributed by
# Rituraj JainC#
// C# program to implement
// the above approach
using System;
class GFG
{
// Function that returns the Kth set bit
static int FindIndexKthBit(int n, int k)
{
int cnt = 0;
int ind = 0;
// Traverse in the binary
while (n > 0)
{
// Check if the last
// bit is set or not
if ((n & 1 ) != 0)
cnt++;
// Check if count is equal to k
// then return the index
if (cnt == k)
return ind;
// Increase the index
// as we move right
ind++;
// Right shift the number by 1
n = n >> 1;
}
return -1;
}
// Driver Code
public static void Main()
{
int n = 15, k = 3;
int ans = FindIndexKthBit(n, k);
if (ans != -1)
Console.WriteLine(ans);
else
Console.WriteLine("No k-th set bit");
}
}
// This code is contributed by
// Code_Mech.PHP
> 1;
}
return -1;
}
// Driver Code
$n = 15;
$k = 3;
$ans = FindIndexKthBit($n, $k);
if ($ans != -1)
echo $ans;
else
echo "No k-th set bit";
// This code is contributed by Ryuga
?>Javascript
输出：
2