https://www.nowcoder.com/practice/8ee967e43c2c4ec193b040ea7fbb10b8
这道题思路不难，关键考察如何将十进制转为二进制。

除二取余

复习了一下如何“除二取余”运算，但明显这种方法比较臃肿了。
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  1. public static int NumberOf1(int n) {
  2.     int binary_n;
  3.     // 如果是负数，需要加上2^31次方
  4.     if (n < 0) {
  5.         binary_n = (int) (Math.pow(2, 31) + n);
  6.     } else {
  7.         binary_n = n;
  8.     }
  10.     String strBinary = "";
  11.     // 通过除二取余法，求出二进制数，保存在binary中
  12.     int shang = binary_n;
  13.     int yu;
  14.     int bit = 0;
  15.     while (shang != 0) {
  16.         yu = shang % 2;
  17.         shang = shang / 2;
  18.         bit++;
  19.         if (bit % 4 == 0) {
  20.             strBinary = " " + yu + strBinary;
  21.         } else {
  22.             strBinary = yu + strBinary;
  23.         }
  24.     }
  26.     // 不够32位，需要补齐
  27.     while (bit < 31) {
  28.         bit++;
  29.         if (bit % 4 == 0) {
  30.             strBinary = " " + "0" + strBinary;
  31.         } else {
  32.             strBinary = "0" + strBinary;
  33.         }
  34.     }
  36.     //判断符号位
  37.     if (n < 0) {
  38.         strBinary = "1" + strBinary;
  39.     } else {
  40.         strBinary = "0" + strBinary;
  41.     }
  43.     System.out.println(strBinary);
  45.     int count = 0;
  46.     for (char c : strBinary.toCharArray()) {
  47.         if (c == '1')
  48.         count++;
  49.     }
  50.     return count;
  51. }