华为OD机考题(HJ62 查找输入整数二进制中1的个数)

前言

经过前期的数据结构和算法学习，开始以OD机考题作为练习题，继续加强下熟练程度。

描述

输入一个正整数，计算它在二进制下的1的个数。

注意多组输入输出！！！！！！

数据范围： 1≤𝑛≤231−1 1≤n≤231−1

输入描述：

输入一个整数

输出描述：

计算整数二进制中1的个数

示例1

输入：

5

输出：

2

说明：

5的二进制表示是101，有2个1

实现原理与步骤

最简单的可以通过Java自带API实现

实现代码(API)

public class CountBits {
    public static void main(String[] args) {
        int number = 29; // Example number
        int count = Integer.bitCount(number);
        System.out.println("Number of 1s in binary representation of " + number + " is: " + count);
    }
}

实现代码(逐位检查)

public class CountBits {
    public static void main(String[] args) {
        int number = 29; // Example number
        int count = countBits(number);
        System.out.println("Number of 1s in binary representation of " + number + " is: " + count);
    }

    public static int countBits(int number) {
        int count = 0;
        while (number != 0) {
            count += number & 1; // Check the least significant bit
            number >>= 1; // Right shift by 1
        }
        return count;
    }
}

实现代码(位清零算法)

public class CountBits {
    public static void main(String[] args) {
        int number = 29; // Example number
        int count = countBits(number);
        System.out.println("Number of 1s in binary representation of " + number + " is: " + count);
    }

    public static int countBits(int number) {
        int count = 0;
        while (number != 0) {
            number &= (number - 1); // Clear the least significant bit set
            count++;
        }
        return count;
    }
}

实现代码(递归算法)

public class CountBits {
    public static void main(String[] args) {
        int number = 29; // Example number
        int count = countBits(number);
        System.out.println("Number of 1s in binary representation of " + number + " is: " + count);
    }

    public static int countBits(int number) {
        if (number == 0) {
            return 0;
        } else {
            return (number & 1) + countBits(number >> 1);
        }
    }
}