给定数字N,请打印所有数字,这些数字是N的二进制表示的按位与集合。数字N的按位与集合是所有可能的数字x小于或等于N,使得对于某些情况,N&i等于x第一号
例子 :
Input : N = 5
Output : 0, 1, 4, 5
Explanation: 0 & 5 = 0
1 & 5 = 1
2 & 5 = 0
3 & 5 = 1
4 & 5 = 4
5 & 5 = 5
So we get 0, 1, 4 and 5 in the
bitwise subsets of N.
Input : N = 9
Output : 0, 1, 8, 9
简单方法:天真的方法是从0到N的所有数字进行迭代,并检查是否(N&i == i)。打印满足指定条件的数字。
下面是上述想法的实现:
C++
// CPP program to print all bitwise
// subsets of N (Naive approach)
#include
using namespace std;
// function to find bitwise subsets
// Naive approach
void printSubsets(int n) {
for (int i = 0; i <= n; i++)
if ((n & i) == i)
cout << i << " ";
}
// Driver Code
int main() {
int n = 9;
printSubsets(n);
return 0;
}
Java
// JAVA program to print all bitwise
// subsets of N (Naive approach)
class GFG {
// function to find bitwise subsets
// Naive approach
static void printSubsets(int n)
{
for (int i = 0; i <= n; i++)
if ((n & i) == i)
System.out.print(i + " ");
}
// Driver function
public static void main(String[] args)
{
int n = 9;
printSubsets(n);
}
}
// This code is contributed by Anant Agarwal.
Python3
# Python program to print all bitwise
# subsets of N (Naive approach)
def printSubsets(n):
for i in range(n + 1):
if ((n & i) == i):
print(i ," ", end = "")
# Driver code
n = 9
printSubsets(n)
# This code is contributed by Anant Agarwal.
C#
// C# program to print all bitwise
// subsets of N (Naive approach)
using System;
class GFG {
// function to find bitwise subsets
// Naive approach
static void printSubsets(int n)
{
for (int i = 0; i <= n; i++)
if ((n & i) == i)
Console.Write(i + " ");
}
// Driver function
public static void Main()
{
int n = 9;
printSubsets(n);
}
}
// This code is contributed by vt_m.
PHP
Javascript
C++
// CPP program to print all bitwise
// subsets of N (Efficient approach)
#include
using namespace std;
// function to find bitwise subsets
// Efficient approach
void printSubsets(int n) {
for (int i = n; i > 0; i = (i - 1) & n)
cout << i << " ";
cout << 0;
}
// Driver Code
int main() {
int n = 9;
printSubsets(n);
return 0;
}
Java
// Java program to print all bitwise
// subsets of N (Efficient approach)
class GFG
{
// function to find bitwise
// subsets Efficient approach
static void printSubsets(int n)
{
for (int i = n; i > 0; i = (i - 1) & n)
System.out.print(i + " ");
System.out.print(" 0 ");
}
// Driver Code
public static void main(String[] args)
{
int n = 9;
printSubsets(n);
}
}
// This code is contributed by ajit.
Python3
# Python 3 program to
# print all bitwise
# subsets of N
# (Efficient approach)
# function to find
# bitwise subsets
# Efficient approach
def printSubsets(n):
i=n
while(i != 0):
print(i,end=" ")
i=(i - 1) & n
print("0")
# Driver Code
n = 9
printSubsets(n)
# This code is contributed by
# Smith Dinesh Semwal
C#
// C# program to print all bitwise
// subsets of N (Efficient approach)
using System;
public class GFG {
// function to find bitwise subsets
// Efficient approach
static void printSubsets(int n) {
for (int i = n; i > 0; i = (i - 1) & n)
Console.Write(i +" ");
Console.WriteLine("0");
}
// Driver Code
static public void Main () {
int n = 9;
printSubsets(n);
}
}
// This code is contributed by vt_m.
PHP
0;
$i = ($i - 1) & $n)
echo $i." ";
echo "0";
}
// Driver Code
$n = 9;
printSubsets($n);
// This code is contributed by mits
?>
Javascript
输出 :
0 1 8 9
时间复杂度: O(N)
高效的解决方案:高效的解决方案是使用按位运算运算符查找子集。除了对每个i进行迭代,我们还可以仅对按位子集进行迭代。向后迭代i =(i-1)&n给我们每个按位子集,其中i从n开始并以1结尾。
下面是上述想法的实现:
C++
// CPP program to print all bitwise
// subsets of N (Efficient approach)
#include
using namespace std;
// function to find bitwise subsets
// Efficient approach
void printSubsets(int n) {
for (int i = n; i > 0; i = (i - 1) & n)
cout << i << " ";
cout << 0;
}
// Driver Code
int main() {
int n = 9;
printSubsets(n);
return 0;
}
Java
// Java program to print all bitwise
// subsets of N (Efficient approach)
class GFG
{
// function to find bitwise
// subsets Efficient approach
static void printSubsets(int n)
{
for (int i = n; i > 0; i = (i - 1) & n)
System.out.print(i + " ");
System.out.print(" 0 ");
}
// Driver Code
public static void main(String[] args)
{
int n = 9;
printSubsets(n);
}
}
// This code is contributed by ajit.
Python3
# Python 3 program to
# print all bitwise
# subsets of N
# (Efficient approach)
# function to find
# bitwise subsets
# Efficient approach
def printSubsets(n):
i=n
while(i != 0):
print(i,end=" ")
i=(i - 1) & n
print("0")
# Driver Code
n = 9
printSubsets(n)
# This code is contributed by
# Smith Dinesh Semwal
C#
// C# program to print all bitwise
// subsets of N (Efficient approach)
using System;
public class GFG {
// function to find bitwise subsets
// Efficient approach
static void printSubsets(int n) {
for (int i = n; i > 0; i = (i - 1) & n)
Console.Write(i +" ");
Console.WriteLine("0");
}
// Driver Code
static public void Main () {
int n = 9;
printSubsets(n);
}
}
// This code is contributed by vt_m.
的PHP
0;
$i = ($i - 1) & $n)
echo $i." ";
echo "0";
}
// Driver Code
$n = 9;
printSubsets($n);
// This code is contributed by mits
?>
Java脚本
输出 :
9 8 1 0
时间复杂度:O(K),其中K是N的按位子集的数量。