首页 > 代码库 > 338. 1的位数 Counting Bits

338. 1的位数 Counting Bits

Given a non negative integer number num. For every numbers i in the range 0 ≤ i ≤ num calculate the number of 1‘s in their binary representation and return them as an array.

Example:
For num = 5 you should return [0,1,1,2,1,2].

Follow up:

  • It is very easy to come up with a solution with run time O(n*sizeof(integer)). But can you do it in linear time O(n) /possibly in a single pass?
  • Space complexity should be O(n).
  • Can you do it like a boss? Do it without using any builtin function like __builtin_popcount in c++ or in any other language.

Hint:

  1. You should make use of what you have produced already.
  2. Divide the numbers in ranges like [2-3], [4-7], [8-15] and so on. And try to generate new range from previous.

技术分享


规律:

i等于二的N次幂时,Hamming weight为 1.

HammingWeight(2) = 1

HammingWeight(3) = HammingWeight(2) + HammingWeight(1) = 2

......

HammingWeight(8) = 1

HammingWeight(9) = HammingWeight(8) + HammingWeight(1) = 2

HammingWeight(10) = HammingWeight(8) + HammingWeight(2) = 2

HammingWeight(11) = HammingWeight(8) + HammingWeight(2) = 3


  1. static public int[] CountBits(int n)
  2. {
  3. int[] arr = new int[n + 1];
  4. arr[0] = 0;
  5. int pow2 = 2;
  6. int before = 1;
  7. for (int i = 1; i < n + 1; i++)
  8. {
  9. if (i == pow2)
  10. {
  11. arr[i] = 1;
  12. before = 1;
  13. pow2 <<= 1;
  14. }
  15. else
  16. {
  17. arr[i] = arr[before] + 1;
  18. before++;
  19. }
  20. }
  21. return arr;
  22. }



来自为知笔记(Wiz)


338. 1的位数 Counting Bits