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.
思路：数字总比[数字二进制最高位置0后的数字]的多一个1
public static int[] countBits(int num) { int len = num + 1; int[] res = new int[len]; res[0] = 0; for (int i = 1; i < len; i++) { res[i] = res[Util.highBit(i)] + 1; } return res; } //将二进制的最高位置0 public static int highBit(int x) { return x - (int) Math.pow(2, (int) (Math.log(x) / Math.log(2))); }
原文：http://www.cnblogs.com/zhaihua/p/5551684.html