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 1:
Input: 2 Output: [0,1,1]Example 2:
Input: 5 Output: [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.这题妙在
十进制： 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16...
二进制中1的个数：0 1 1 2 1 2 2 3 1 2 2 3 2 3 3 4 1...
这题最妙在于当遇到2的pow时，把t置零，然后在前面数字二进制中1的个数的基础上加1
public class Solution { public int[] countBits(int num) { int[] res = new int[num+1]; int pow = 1; res[0] = 0; for(int i=1,t=0;i<=num;i++,t++) { if(i==pow) { pow = pow * 2; t = 0; } res[i] = res[t] + 1; } return res; } }
转载于:https://www.cnblogs.com/hygeia/p/9758094.html
