Early in the morning, a few small problems

Title: Please implement a function, an integer input, the output binary number represents the number 1. For example, to 9 is expressed as a binary 1001, there is a 2 1. Thus, if the input 9, the function output 2.

public class Solution {
    public int NumberOf(int num) {
        int count = 0;
        while (num != 0) {
            if ((num & 1) == 1) {
                count++;
            }
            num = num >>> 1;    //如果使用>>有符号右移，那么符号位1永远会存在，也就是会产生死循环
        }
    }
}
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Implement a function Math.pow, the m-th power of n -.

public class Solution {
    public double pow(int m, int n) {
        double result = 0;
        if (m == 0 && n < 0) {
            return -1;
        }
        int absN = Math.abs(n);
        result = calc(m, absN);
        if (n < 0) {
            result = 1 / result;
        }
        return result;
    }
    
    private int calc(int m, int n) {
        int result = 1;
        for (int i = 0; i < n; i++) {
            result *= m;
        }
        return result;
    }
}
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Improved calc

private int calc(int m, int n) {
   if (n == 0) {
       return 1;
   }
   if (n == 1) {
       return m;
   }
   int result = calc(m, n >> 1);    //右移1位表示除以2
   result *= result;
   if ((m & 1) == 1) {     //位运算判断是会否为奇数，奇数的二进制第一位一定是1与1做与运算即可判断是否为奇数，代替m%2是否等于0
       result *= m;
   }
   return result;
}
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Short step a thousand miles.

Reproduced in: https: //juejin.im/post/5d0c1fc4f265da1bbd4b7c61