C程序以数字的二进制表示形式对零和一进行计数

给定数字N ，任务是编写C程序以N的二进制表示形式对0和1的数目进行计数。

例子：

Input: N = 5 
Output: 
Count of 0s: 1 
Count of 1s: 2 
Explanation: Binary representation of 5 is “101”.
Input: N = 22 
Output: 
Count of 0s: 2 
Count of 1s: 3 
Explanation: Binary representation of 22 is “10110”. 

方法1-天真的方法：想法是遍历N的二进制表示形式中的所有位，如果当前位为‘0’ ，则递增0s的计数，否则递增1s的计数。

下面是上述方法的实现：

C

// C program for the above approach
#include 
 
// Function to count the number of 0s
// and 1s in binary representation of N
void count1s0s(int N)
{
    // Initialise count variables
    int count0 = 0, count1 = 0;
 
    // Iterate through all the bits
    while (N > 0) {
 
        // If current bit is 1
        if (N & 1) {
            count1++;
        }
 
        // If current bit is 0
        else {
            count0++;
        }
 
        N = N >> 1;
    }
 
    // Print the count
    printf("Count of 0s in N is %d\n", count0);
    printf("Count of 1s in N is %d\n", count1);
}
 
// Driver Code
int main()
{
    // Given Number
    int N = 9;
 
    // Function Call
    count1s0s(N);
    return 0;
}

输出

Count of 0s in N is 2
Count of 1s in N is 2

时间复杂度： O(log N)

方法2 –递归方法：上述方法也可以使用递归来实现。

下面是上述方法的实现：

C

// C program for the above approach
#include 
#include 
 
// Recursive approach to find the
// number of set bit in 1
int recursiveCount(int N)
{
 
    // Base Case
    if (N == 0) {
        return 0;
    }
 
    // Return recursively
    return (N & 1) + recursiveCount(N >> 1);
}
 
// Function to find 1s complement
int onesComplement(int n)
{
    // Find number of bits in the
    // given integer
    int N = floor(log2(n)) + 1;
 
    // XOR the given integer with
    // pow(2, N) - 1
    return ((1 << N) - 1) ^ n;
}
 
// Function to count the number of 0s
// and 1s in binary representation of N
void count1s0s(int N)
{
    // Initialise the count variables
    int count0, count1;
 
    // Function call to find the number
    // of set bits in N
    count1 = recursiveCount(N);
 
    // Function call to find 1s complement
    N = onesComplement(N);
 
    // Function call to find the number
    // of set bits in 1s complement of N
    count0 = recursiveCount(N);
 
    // Print the count
    printf("Count of 0s in N is %d\n", count0);
    printf("Count of 1s in N is %d\n", count1);
}
 
// Driver Code
int main()
{
    // Given Number
    int N = 5;
 
    // Function Call
    count1s0s(N);
    return 0;
}

输出

Count of 0s in N is 1
Count of 1s in N is 2

时间复杂度： O(log N)

方法3 –使用Brian Kernighan的算法
我们可以使用以下步骤找到设置位的数量：

• 初始化计数为0 。
• 如果N> 0 ，则将N更新为N＆(N – 1)，因为这将从右侧取消设置最大的位，如下所示：

if N = 10;
Binary representation of N     = 1010
Binary representation of N - 1 = 1001
-------------------------------------
Logical AND of N and N - 1     = 1000

• 增加以上步骤的计数，并重复以上步骤，直到l N变为0。

要在N的二进制表示形式中找到0的计数，请使用上述方法找到N的一个补数并找到置位的计数。

下面是上述方法的实现：

C

// C program for the above approach
#include 
#include 
 
// Function to find 1s complement
int onesComplement(int n)
{
    // Find number of bits in the
    // given integer
    int N = floor(log2(n)) + 1;
 
    // XOR the given integer with
    // pow(2, N) - 1
    return ((1 << N) - 1) ^ n;
}
 
// Function to implement count of
// set bits using Brian Kernighan’s
// Algorithm
int countSetBits(int n)
{
    // Initialise count
    int count = 0;
 
    // Iterate untill n is 0
    while (n) {
        n &= (n - 1);
        count++;
    }
 
    // Return the final count
    return count;
}
 
// Function to count the number of 0s
// and 1s in binary representation of N
void count1s0s(int N)
{
    // Initialise the count variables
    int count0, count1;
 
    // Function call to find the number
    // of set bits in N
    count1 = countSetBits(N);
 
    // Function call to find 1s complement
    N = onesComplement(N);
 
    // Function call to find the number
    // of set bits in 1s complement of N
    count0 = countSetBits(N);
 
    // Print the count
    printf("Count of 0s in N is %d\n", count0);
    printf("Count of 1s in N is %d\n", count1);
}
 
// Driver Code
int main()
{
    // Given Number
    int N = 5;
 
    // Function Call
    count1s0s(N);
    return 0;
}

输出

Count of 0s in N is 1
Count of 1s in N is 2

时间复杂度： O(log N)

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