给定一个非负整数N ,任务是找到从0到N 的每个数字的设置位数。
例子:
Input: N = 3
Output: 0 1 1 2
0, 1, 2 and 3 can be written in binary as 0, 1, 10 and 11.
The number of 1’s in their binary representation are 0, 1, 1 and 2.
Input: N = 5
Output: 0 1 1 2 1 2
天真的方法:运行从 0 到 N 的循环,并使用内置的位计数函数__builtin_popcount(),找到所有所需整数中的设置位数。
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to find the count
// of set bits in all the
// integers from 0 to n
void findSetBits(int n)
{
for (int i = 0; i <= n; i++)
cout << __builtin_popcount(i) << " ";
}
// Driver code
int main()
{
int n = 5;
findSetBits(n);
return 0;
}
Java
// Java implementation of the approach
class GFG
{
// Function to find the count
// of set bits in all the
// integers from 0 to n
static void findSetBits(int n)
{
for (int i = 0; i <= n; i++)
System.out.print(Integer.bitCount(i) + " ");
}
// Driver code
public static void main(String[] args)
{
int n = 5;
findSetBits(n);
}
}
// This code is contributed by Rajput-Ji
Python 3
# Python 3 implementation of the approach
def count(n):
count = 0
while (n):
count += n & 1
n >>= 1
return count
# Function to find the count
# of set bits in all the
# integers from 0 to n
def findSetBits(n):
for i in range(n + 1):
print(count(i), end = " ")
# Driver code
if __name__ == '__main__':
n = 5
findSetBits(n)
# This code is contributed by Surendra_Gangwar
C#
// C# implementation of the approach
using System;
class GFG
{
static int count(int n)
{
int count = 0;
while (n > 0)
{
count += n & 1;
n >>= 1;
}
return count;
}
// Function to find the count
// of set bits in all the
// integers from 0 to n
static void findSetBits(int n)
{
for (int i = 0; i <= n; i++)
Console.Write(count(i)+" ");
}
// Driver code
public static void Main(String []args)
{
int n = 5;
findSetBits(n);
}
}
// This code is contributed by SHUBHAMSINGH10
Javascript
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to find the count
// of set bits in all the
// integers from 0 to n
void findSetBits(int n)
{
// dp[i] will store the count
// of set bits in i
int dp[n + 1];
// Initialise the dp array
memset(dp, 0, sizeof(dp));
// Count of set bits in 0 is 0
cout << dp[0] << " ";
// For every number starting from 1
for (int i = 1; i <= n; i++) {
// If current number is even
if (i % 2 == 0) {
// Count of set bits in i is equal to
// the count of set bits in (i / 2)
dp[i] = dp[i / 2];
}
// If current element is odd
else {
// Count of set bits in i is equal to
// the count of set bits in (i / 2) + 1
dp[i] = dp[i / 2] + 1;
}
// Print the count of set bits in i
cout << dp[i] << " ";
}
}
// Driver code
int main()
{
int n = 5;
findSetBits(n);
return 0;
}
Java
// Java implementation of the approach
class GFG
{
// Function to find the count
// of set bits in all the
// integers from 0 to n
static void findSetBits(int n)
{
// dp[i] will store the count
// of set bits in i
int []dp = new int[n + 1];
// Count of set bits in 0 is 0
System.out.print(dp[0] + " ");
// For every number starting from 1
for (int i = 1; i <= n; i++)
{
// If current number is even
if (i % 2 == 0)
{
// Count of set bits in i is equal to
// the count of set bits in (i / 2)
dp[i] = dp[i / 2];
}
// If current element is odd
else
{
// Count of set bits in i is equal to
// the count of set bits in (i / 2) + 1
dp[i] = dp[i / 2] + 1;
}
// Print the count of set bits in i
System.out.print(dp[i] + " ");
}
}
// Driver code
public static void main(String []args)
{
int n = 5;
findSetBits(n);
}
}
// This code is contributed by Rajput-Ji
Python3
# Python3 implementation of the approach
# Function to find the count of set bits
# in all the integers from 0 to n
def findSetBits(n) :
# dp[i] will store the count
# of set bits in i
# Initialise the dp array
dp = [0] * (n + 1);
# Count of set bits in 0 is 0
print(dp[0], end = " ");
# For every number starting from 1
for i in range(1, n + 1) :
# If current number is even
if (i % 2 == 0) :
# Count of set bits in i is equal to
# the count of set bits in (i / 2)
dp[i] = dp[i // 2];
# If current element is odd
else :
# Count of set bits in i is equal to
# the count of set bits in (i / 2) + 1
dp[i] = dp[i // 2] + 1;
# Print the count of set bits in i
print(dp[i], end = " ");
# Driver code
if __name__ == "__main__" :
n = 5;
findSetBits(n);
# This code is contributed by AnkitRai01
C#
// C# implementation of the approach
using System;
class GFG
{
// Function to find the count
// of set bits in all the
// integers from 0 to n
static void findSetBits(int n)
{
// dp[i] will store the count
// of set bits in i
int []dp = new int[n + 1];
// Count of set bits in 0 is 0
Console.Write(dp[0] + " ");
// For every number starting from 1
for (int i = 1; i <= n; i++)
{
// If current number is even
if (i % 2 == 0)
{
// Count of set bits in i is equal to
// the count of set bits in (i / 2)
dp[i] = dp[i / 2];
}
// If current element is odd
else
{
// Count of set bits in i is equal to
// the count of set bits in (i / 2) + 1
dp[i] = dp[i / 2] + 1;
}
// Print the count of set bits in i
Console.Write(dp[i] + " ");
}
}
// Driver code
public static void Main(String []args)
{
int n = 5;
findSetBits(n);
}
}
// This code is contributed by 29AjayKumar
Javascript
0 1 1 2 1 2
有效的方法:让我们写出范围 (0, 6) 内数字的二进制表示。
0 in binary – 000
1 in binary – 001
2 in binary – 010
3 in binary – 011
4 in binary – 100
5 in binary – 101
6 in binary – 110
因为,任何偶数都可以写成(2 * i) ,任何奇数都可以写成(2 * i + 1) ,其中i是自然数。
2, 4和3, 6在其二进制表示中具有相同数量的 1,因为乘以任何数字相当于将其左移 1(阅读此处)。
类似地,任何偶数2 * i和i在其二进制表示中将具有相同数量的1 。
1个在5(101)的数目为2的二进制表示+ 1等于1倍的数。因此,对于任何奇数(2 * i + 1) ,它将是 ( i的二进制表示中1 的数量) + 1 。
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to find the count
// of set bits in all the
// integers from 0 to n
void findSetBits(int n)
{
// dp[i] will store the count
// of set bits in i
int dp[n + 1];
// Initialise the dp array
memset(dp, 0, sizeof(dp));
// Count of set bits in 0 is 0
cout << dp[0] << " ";
// For every number starting from 1
for (int i = 1; i <= n; i++) {
// If current number is even
if (i % 2 == 0) {
// Count of set bits in i is equal to
// the count of set bits in (i / 2)
dp[i] = dp[i / 2];
}
// If current element is odd
else {
// Count of set bits in i is equal to
// the count of set bits in (i / 2) + 1
dp[i] = dp[i / 2] + 1;
}
// Print the count of set bits in i
cout << dp[i] << " ";
}
}
// Driver code
int main()
{
int n = 5;
findSetBits(n);
return 0;
}
Java
// Java implementation of the approach
class GFG
{
// Function to find the count
// of set bits in all the
// integers from 0 to n
static void findSetBits(int n)
{
// dp[i] will store the count
// of set bits in i
int []dp = new int[n + 1];
// Count of set bits in 0 is 0
System.out.print(dp[0] + " ");
// For every number starting from 1
for (int i = 1; i <= n; i++)
{
// If current number is even
if (i % 2 == 0)
{
// Count of set bits in i is equal to
// the count of set bits in (i / 2)
dp[i] = dp[i / 2];
}
// If current element is odd
else
{
// Count of set bits in i is equal to
// the count of set bits in (i / 2) + 1
dp[i] = dp[i / 2] + 1;
}
// Print the count of set bits in i
System.out.print(dp[i] + " ");
}
}
// Driver code
public static void main(String []args)
{
int n = 5;
findSetBits(n);
}
}
// This code is contributed by Rajput-Ji
蟒蛇3
# Python3 implementation of the approach
# Function to find the count of set bits
# in all the integers from 0 to n
def findSetBits(n) :
# dp[i] will store the count
# of set bits in i
# Initialise the dp array
dp = [0] * (n + 1);
# Count of set bits in 0 is 0
print(dp[0], end = " ");
# For every number starting from 1
for i in range(1, n + 1) :
# If current number is even
if (i % 2 == 0) :
# Count of set bits in i is equal to
# the count of set bits in (i / 2)
dp[i] = dp[i // 2];
# If current element is odd
else :
# Count of set bits in i is equal to
# the count of set bits in (i / 2) + 1
dp[i] = dp[i // 2] + 1;
# Print the count of set bits in i
print(dp[i], end = " ");
# Driver code
if __name__ == "__main__" :
n = 5;
findSetBits(n);
# This code is contributed by AnkitRai01
C#
// C# implementation of the approach
using System;
class GFG
{
// Function to find the count
// of set bits in all the
// integers from 0 to n
static void findSetBits(int n)
{
// dp[i] will store the count
// of set bits in i
int []dp = new int[n + 1];
// Count of set bits in 0 is 0
Console.Write(dp[0] + " ");
// For every number starting from 1
for (int i = 1; i <= n; i++)
{
// If current number is even
if (i % 2 == 0)
{
// Count of set bits in i is equal to
// the count of set bits in (i / 2)
dp[i] = dp[i / 2];
}
// If current element is odd
else
{
// Count of set bits in i is equal to
// the count of set bits in (i / 2) + 1
dp[i] = dp[i / 2] + 1;
}
// Print the count of set bits in i
Console.Write(dp[i] + " ");
}
}
// Driver code
public static void Main(String []args)
{
int n = 5;
findSetBits(n);
}
}
// This code is contributed by 29AjayKumar
Javascript
0 1 1 2 1 2
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